Evaluate the intgral $\int{\frac{dx}{x^2(1-x^2)}}$ (solution verification) I have to solve the following integral
$$\int{\frac{dx}{x^2(1-x^2)}}$$
What I've got:
\begin{split}
\int{\frac{dx}{x^2(1-x^2)}} &=\int{\frac{(1-x^2+x^2)dx}{x^2(1-x^2)}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{1-x^2}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{(1+(xi)^2)}}\\
&=-x^{-1}+\arctan{xi}+C
\end{split}
Is this correct?
Thanks in advance!

Note that, if $f(x)=\arctan(xi)$, then
$$
f'(x)=i\frac{1}{1+(xi)^2}=\frac{i}{1-x^2}
$$
that's not the required derivative; you are missing $1/i=-i$, so you could write
$$
-\frac{1}{x}-i\arctan(xi)
$$

I'd prefer to avoid complex functions. With partial fractions, we set
$$
\frac{1}{x^2(1-x)^2}=
\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x-1}+\frac{d}{x+1}
$$
Removing the denominators leads to
$$
ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
$$
With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.
Therefore
$$
\int\frac{1}{x^2(1-x^2)}\,dx=
\int\left(
  \frac{1}{x^2}-\frac{1}{2}\frac{1}{x-1}+\frac{1}{2}\frac{1}{x+1}
\right)\,dx=-\frac{1}{x}+\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|+c
$$
