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How to find the antiderivative of $\dfrac{1}{x^2(1+x^2)}$? I recognized that this can be done with trigonometric substitution and I let $x = \tan(x)$ and ended up with $\dfrac{1}{\tan(x^2)}$; then I got stuck with it.
I have the answer for this question, which is $-\arctan(x) - \dfrac{1}{x} + C$. However, I would really want to learn how it is solved.

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    $\begingroup$ It will be very easy to get confused if you don't change variable names when making a substitution - use something like x = tan(u) rather than x=tan(x). $\endgroup$ – Mark Pattison Aug 8 '14 at 13:42
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Hint:$$\frac{1}{x^2(x^2+1)}=\frac{1}{x^2}-\frac{1}{x^2+1}$$

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  • $\begingroup$ helpful, thanks! $\endgroup$ – Joshua L Aug 8 '14 at 12:09
  • $\begingroup$ Effective and helpful! :) $\endgroup$ – mrs Aug 8 '14 at 12:09
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As an alternative to partial fractions, you could try using the substitution $u=\frac{1}{x}$. Then,

$$du=-\frac{dx}{x^2}$$

and

$$\frac{1}{1+x^2}=\frac{1}{1+\frac{1}{u^2}}=\frac{u^2}{u^2+1}=1-\frac{1}{u^2+1}.$$

Then the anti-derivative sought follows immediately.

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If you want to go on your way, after getting $$\int\frac{du}{\tan^2 u}$$ (note that it is not $\int du/\tan(u^2)$) by using $x=\tan u$, you can use $$\frac{1}{\tan^2u}=\frac{1}{\sin^2u}-1,\ \ \left(-\frac{1}{\tan u}\right)'=\frac{1}{\sin^2u}.$$

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