How to find the antiderivative of $\frac{1}{x^2(1+x^2)}$? 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.

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.
