find the sum of $ \sum_{n=1}^{\infty} \frac{x^{2n}(-1)^n}{n} $ $ \sum_{n=1}^{\infty} \frac{x^{2n}(-1)^n}{n} $
compute the sum of this
$\frac{1}{1+x^2}$ = $ \sum_{n=1}^{\infty} x^{2n}(-1)^n $
and then integrate both sides 
$ \sum_{n=1}^{\infty} \frac{x^{2n+1}(-1)^n}{2n+1} $
but how to get n from there

You know that:
$\begin{align}
\frac{1}{1 - z}
  &= \sum_{n \ge 0} z^n \\
\int_0^z \frac{\mathrm{d} t}{1 - t}
  &= - \ln (1 - z) \\
  &= \sum_{n \ge 0} \frac{z^{n + 1}}{n + 1}
\end{align}$
so that:
$\begin{align}
\sum_{n \ge 1} \frac{(-1)^n x^{2 n}}{n}
  &= - \ln (1 + x^2)
\end{align}$
