Expressing $\displaystyle\int_1^x S(u) \mathrm{d}u$ in terms of elementary function Let's suppose we have the following series 
$$S(x)=\sum_{n=1}^{\infty}(-1)^n\frac{x^{2n}}{n+1}$$
then I need to express $\displaystyle\int_1^x S(u) \mathrm{d}u$ in terms of elementary function. Some hints, suggestions ?
Thanks!!!
Sis.
 A: We have 
$$\log(1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n},$$
so we find
$$x^2S(x)=\sum_{n=1}^\infty(-1)^n\frac{(x^2)^{n+1}}{n+1}=\sum_{n=1}^\infty(-1)^{n+1}\frac{(x^2)^n}{n}-x^2=\log(1+x^2)-x^2.$$
Now, we inegrate by parts, we find
\begin{align}\int_1^xS(u)du&=\int_1^x(\frac{\log(1+u^2)}{u^2}-1)du\\&=\frac{-1}{2x}(2\log(x^2+1)-(2-\pi+2\log2)x-4x\arctan x+2x^2)\end{align}
A: There are two elementary approaches, recognize and integrate, or integrate and recognize.  We take the second path. 
Your sum goes from $n=1$ to infinity, but if it is a good little mathematical sum, it probably really wants to start at $n=0$. If it stubbornly insists on $n=1$, we can subtract appropriately at the end.  When we integrate
$$\sum_0^\infty (-1)^n \frac{x^{2n}}{n+1}$$
term by term, we get
$$\sum_0^\infty (-1)^n \frac{x^{2n+1}}{(n+1)(2n+1)},$$
or equivalently (partial fractions)
$$\sum_0^\infty 2(-1)^n\frac{x^{2n+1}}{2n+1} - \sum_0^\infty (-1)^n \frac{x^{2n+1}}{n+1}.$$
The first sum is $2\arctan x$.  The second sum (for $x\ne 0$) can be rewritten as 
$\dfrac{1}{x}\displaystyle\sum_0^\infty (-1)^n \dfrac{(x^2)^{n+1}}{n+1}$. This is $\dfrac{1}{x}\log(1+x^2)$. 
