Teaching indefinite integrals that require special-casing












12












$begingroup$


I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?



Let's consider the following example.



Find the indefinite integral
$$
I=intdfrac{dx}{xsqrt{x^{2}-1}}.
$$

Some of my students gave the following answer.



Let $t=1/x$ then $dx=-1/t^{2}dt$, so we get
$$
I=intdfrac{-1/t^{2}dt}{frac{1}{t}sqrt{frac{1}{t^{2}}-1}}=intdfrac{-dt}{sqrt{1-t^{2}}}=-arcsinleft(tright)+C=-arcsinleft(frac{1}{x}right)+C.
$$



Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac{-1/t^{2}dt}{frac{1}{t}sqrt{frac{1}{t^{2}}-1}}=intdfrac{-left|tright|dt}{tsqrt{1-t^{2}}}.
$$

Then we end up with the answer
$$
intdfrac{dx}{xsqrt{x^{2}-1}}=begin{cases}
-arcsinleft(dfrac{1}{x}right)+C & text{for }x>1,\
arcsinleft(dfrac{1}{x}right)+C & text{for }x<-1.
end{cases}
$$

In your teaching practice, how would you usually proceed?



PS. We may encounter the same issue in many other problems. For example, find $intsqrt{1-x^{2}}dx$. Then if we let $x=sinleft(tright)$ then
$sqrt{1-sin^{2}left(tright)}$ should be $left|cosleft(tright)right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.










share|improve this question











$endgroup$








  • 1




    $begingroup$
    Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
    $endgroup$
    – Acccumulation
    22 hours ago






  • 5




    $begingroup$
    The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
    $endgroup$
    – Javier
    20 hours ago










  • $begingroup$
    By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
    $endgroup$
    – kcrisman
    13 hours ago
















12












$begingroup$


I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?



Let's consider the following example.



Find the indefinite integral
$$
I=intdfrac{dx}{xsqrt{x^{2}-1}}.
$$

Some of my students gave the following answer.



Let $t=1/x$ then $dx=-1/t^{2}dt$, so we get
$$
I=intdfrac{-1/t^{2}dt}{frac{1}{t}sqrt{frac{1}{t^{2}}-1}}=intdfrac{-dt}{sqrt{1-t^{2}}}=-arcsinleft(tright)+C=-arcsinleft(frac{1}{x}right)+C.
$$



Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac{-1/t^{2}dt}{frac{1}{t}sqrt{frac{1}{t^{2}}-1}}=intdfrac{-left|tright|dt}{tsqrt{1-t^{2}}}.
$$

Then we end up with the answer
$$
intdfrac{dx}{xsqrt{x^{2}-1}}=begin{cases}
-arcsinleft(dfrac{1}{x}right)+C & text{for }x>1,\
arcsinleft(dfrac{1}{x}right)+C & text{for }x<-1.
end{cases}
$$

In your teaching practice, how would you usually proceed?



PS. We may encounter the same issue in many other problems. For example, find $intsqrt{1-x^{2}}dx$. Then if we let $x=sinleft(tright)$ then
$sqrt{1-sin^{2}left(tright)}$ should be $left|cosleft(tright)right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.










share|improve this question











$endgroup$








  • 1




    $begingroup$
    Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
    $endgroup$
    – Acccumulation
    22 hours ago






  • 5




    $begingroup$
    The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
    $endgroup$
    – Javier
    20 hours ago










  • $begingroup$
    By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
    $endgroup$
    – kcrisman
    13 hours ago














12












12








12


1



$begingroup$


I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?



Let's consider the following example.



Find the indefinite integral
$$
I=intdfrac{dx}{xsqrt{x^{2}-1}}.
$$

Some of my students gave the following answer.



Let $t=1/x$ then $dx=-1/t^{2}dt$, so we get
$$
I=intdfrac{-1/t^{2}dt}{frac{1}{t}sqrt{frac{1}{t^{2}}-1}}=intdfrac{-dt}{sqrt{1-t^{2}}}=-arcsinleft(tright)+C=-arcsinleft(frac{1}{x}right)+C.
$$



Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac{-1/t^{2}dt}{frac{1}{t}sqrt{frac{1}{t^{2}}-1}}=intdfrac{-left|tright|dt}{tsqrt{1-t^{2}}}.
$$

Then we end up with the answer
$$
intdfrac{dx}{xsqrt{x^{2}-1}}=begin{cases}
-arcsinleft(dfrac{1}{x}right)+C & text{for }x>1,\
arcsinleft(dfrac{1}{x}right)+C & text{for }x<-1.
end{cases}
$$

In your teaching practice, how would you usually proceed?



PS. We may encounter the same issue in many other problems. For example, find $intsqrt{1-x^{2}}dx$. Then if we let $x=sinleft(tright)$ then
$sqrt{1-sin^{2}left(tright)}$ should be $left|cosleft(tright)right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.










share|improve this question











$endgroup$




I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?



Let's consider the following example.



Find the indefinite integral
$$
I=intdfrac{dx}{xsqrt{x^{2}-1}}.
$$

Some of my students gave the following answer.



Let $t=1/x$ then $dx=-1/t^{2}dt$, so we get
$$
I=intdfrac{-1/t^{2}dt}{frac{1}{t}sqrt{frac{1}{t^{2}}-1}}=intdfrac{-dt}{sqrt{1-t^{2}}}=-arcsinleft(tright)+C=-arcsinleft(frac{1}{x}right)+C.
$$



Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac{-1/t^{2}dt}{frac{1}{t}sqrt{frac{1}{t^{2}}-1}}=intdfrac{-left|tright|dt}{tsqrt{1-t^{2}}}.
$$

Then we end up with the answer
$$
intdfrac{dx}{xsqrt{x^{2}-1}}=begin{cases}
-arcsinleft(dfrac{1}{x}right)+C & text{for }x>1,\
arcsinleft(dfrac{1}{x}right)+C & text{for }x<-1.
end{cases}
$$

In your teaching practice, how would you usually proceed?



PS. We may encounter the same issue in many other problems. For example, find $intsqrt{1-x^{2}}dx$. Then if we let $x=sinleft(tright)$ then
$sqrt{1-sin^{2}left(tright)}$ should be $left|cosleft(tright)right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.







teaching teacher-preparation






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edited 17 hours ago







Hoa

















asked yesterday









HoaHoa

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1116








  • 1




    $begingroup$
    Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
    $endgroup$
    – Acccumulation
    22 hours ago






  • 5




    $begingroup$
    The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
    $endgroup$
    – Javier
    20 hours ago










  • $begingroup$
    By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
    $endgroup$
    – kcrisman
    13 hours ago














  • 1




    $begingroup$
    Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
    $endgroup$
    – Acccumulation
    22 hours ago






  • 5




    $begingroup$
    The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
    $endgroup$
    – Javier
    20 hours ago










  • $begingroup$
    By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
    $endgroup$
    – kcrisman
    13 hours ago








1




1




$begingroup$
Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
$endgroup$
– Acccumulation
22 hours ago




$begingroup$
Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
$endgroup$
– Acccumulation
22 hours ago




5




5




$begingroup$
The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
$endgroup$
– Javier
20 hours ago




$begingroup$
The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
$endgroup$
– Javier
20 hours ago












$begingroup$
By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
$endgroup$
– kcrisman
13 hours ago




$begingroup$
By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
$endgroup$
– kcrisman
13 hours ago










4 Answers
4






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oldest

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9












$begingroup$

This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.



Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
$$lim_{xtopminfty} frac{x}{sqrt{x^2+1}} = pm 1$$



The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:



$$lim_{xtopminfty} frac{xcdot 1/x}{sqrt{(x^2+1)cdot 1/x^2}} = lim_{xtopminfty} frac{1}{sqrt{1+1/x^2}}=1$$



But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.



In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)



So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac{1}{xsqrt{x^2-1}}$ is definitely negative there.



enter image description here



Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.





Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)






share|improve this answer









$endgroup$





















    8












    $begingroup$

    I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.



    It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)






    share|improve this answer









    $endgroup$





















      5












      $begingroup$

      You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.



      In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!



      The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.






      share|improve this answer









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        4












        $begingroup$

        To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.



        I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.



        As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.






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          4 Answers
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          4 Answers
          4






          active

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          active

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          active

          oldest

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          9












          $begingroup$

          This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.



          Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
          $$lim_{xtopminfty} frac{x}{sqrt{x^2+1}} = pm 1$$



          The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:



          $$lim_{xtopminfty} frac{xcdot 1/x}{sqrt{(x^2+1)cdot 1/x^2}} = lim_{xtopminfty} frac{1}{sqrt{1+1/x^2}}=1$$



          But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.



          In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)



          So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac{1}{xsqrt{x^2-1}}$ is definitely negative there.



          enter image description here



          Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.





          Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)






          share|improve this answer









          $endgroup$


















            9












            $begingroup$

            This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.



            Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
            $$lim_{xtopminfty} frac{x}{sqrt{x^2+1}} = pm 1$$



            The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:



            $$lim_{xtopminfty} frac{xcdot 1/x}{sqrt{(x^2+1)cdot 1/x^2}} = lim_{xtopminfty} frac{1}{sqrt{1+1/x^2}}=1$$



            But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.



            In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)



            So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac{1}{xsqrt{x^2-1}}$ is definitely negative there.



            enter image description here



            Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.





            Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)






            share|improve this answer









            $endgroup$
















              9












              9








              9





              $begingroup$

              This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.



              Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
              $$lim_{xtopminfty} frac{x}{sqrt{x^2+1}} = pm 1$$



              The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:



              $$lim_{xtopminfty} frac{xcdot 1/x}{sqrt{(x^2+1)cdot 1/x^2}} = lim_{xtopminfty} frac{1}{sqrt{1+1/x^2}}=1$$



              But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.



              In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)



              So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac{1}{xsqrt{x^2-1}}$ is definitely negative there.



              enter image description here



              Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.





              Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)






              share|improve this answer









              $endgroup$



              This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.



              Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
              $$lim_{xtopminfty} frac{x}{sqrt{x^2+1}} = pm 1$$



              The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:



              $$lim_{xtopminfty} frac{xcdot 1/x}{sqrt{(x^2+1)cdot 1/x^2}} = lim_{xtopminfty} frac{1}{sqrt{1+1/x^2}}=1$$



              But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.



              In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)



              So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac{1}{xsqrt{x^2-1}}$ is definitely negative there.



              enter image description here



              Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.





              Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered yesterday









              kcrismankcrisman

              3,623731




              3,623731























                  8












                  $begingroup$

                  I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.



                  It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)






                  share|improve this answer









                  $endgroup$


















                    8












                    $begingroup$

                    I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.



                    It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)






                    share|improve this answer









                    $endgroup$
















                      8












                      8








                      8





                      $begingroup$

                      I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.



                      It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)






                      share|improve this answer









                      $endgroup$



                      I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.



                      It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered yesterday









                      Henry TowsnerHenry Towsner

                      7,1102349




                      7,1102349























                          5












                          $begingroup$

                          You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.



                          In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!



                          The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.






                          share|improve this answer









                          $endgroup$


















                            5












                            $begingroup$

                            You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.



                            In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!



                            The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.






                            share|improve this answer









                            $endgroup$
















                              5












                              5








                              5





                              $begingroup$

                              You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.



                              In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!



                              The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.






                              share|improve this answer









                              $endgroup$



                              You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.



                              In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!



                              The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 20 hours ago









                              alephzeroalephzero

                              28113




                              28113























                                  4












                                  $begingroup$

                                  To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.



                                  I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.



                                  As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.






                                  share|improve this answer








                                  New contributor




                                  Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.






                                  $endgroup$


















                                    4












                                    $begingroup$

                                    To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.



                                    I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.



                                    As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.






                                    share|improve this answer








                                    New contributor




                                    Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    $endgroup$
















                                      4












                                      4








                                      4





                                      $begingroup$

                                      To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.



                                      I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.



                                      As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.






                                      share|improve this answer








                                      New contributor




                                      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






                                      $endgroup$



                                      To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.



                                      I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.



                                      As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.







                                      share|improve this answer








                                      New contributor




                                      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      share|improve this answer



                                      share|improve this answer






                                      New contributor




                                      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      answered 17 hours ago









                                      AndrewAndrew

                                      411




                                      411




                                      New contributor




                                      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.





                                      New contributor





                                      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






                                      Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






























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