Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to integrate


what I have try is trigonometric substitution and partition function and integration by part

but still cannot solve it

Thx for your reading!

share|improve this question
Use the substitution y=$1-x²$. Then one finds: $\frac{x^2}{1-x^2}dx=\frac{y'}{4y}dy$. Integration now becomes: $ln(y^{1/4})$. But this seems weird. Is there any error? Thanks for your attention. –  awllower Nov 4 '12 at 9:38
@awllower: It would be $\frac14 \ln y$ if you were integrating with respect to $x$. Integrating $\frac{y'}{y}\ dy$ doesn't make a lot of sense in this context. –  Javier Jan 26 '13 at 2:52
@JavierBadia I could not understand what you claimed here. I meant only to use the theorem of changes of variables in elementary calculus. Why does this make no sense here? Moreover, $(lny)/4=ln(y^{1/4})$, right? –  awllower Jan 30 '13 at 3:51

2 Answers 2

up vote 7 down vote accepted

$$\frac{x^2}{1-x^2} = -\frac{x^2}{x^2 - 1} = -\frac{x^2 - 1 + 1}{x^2 - 1} = -1 - \frac{1}{x^2 - 1}$$ Can you take it from here?

share|improve this answer
that tricky algebra can give students headaches. I always tell my students that if the integrand offers itself up to division then DO IT. Of course it yields the same result that you reached but experience has taught me that saying you are about to divide and then briefly do the long division (of polynomials) students get a quick review of the same and the integral gets easier as a prize. –  DaveUM Jan 26 '13 at 2:37
@DaveUM: Agree. Students usually like algorithms, and rational functions and their relatives give one of the few opportunities. –  André Nicolas Jan 26 '13 at 3:54

And going a little further,

$$\frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} =\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) $$

At this point, even I can integrate it.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.