# How to integrate $\frac{x^2}{1-x^2}$

I want to integrate

$\frac{x^2}{1-x^2},$

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

but still cannot solve it

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 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 Badia Jan 26 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 at 3:51

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