Find $\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$ I want to find the sum:
$$\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$$
I start with finding the antiderivative of the integrant, which is:
$$\frac{1}{2\sqrt{2}}[\ln(x^2-\sqrt{2}x+1)-\ln(x^2+\sqrt{2}x+1)+2\arctan(\sqrt{2}x-1)+2\arctan(\sqrt{2}x+1)]$$
Then I use the fundamental theorem of calculus to evaluate the integral. It turns out that the result is really ugly and I have no idea how to hanled it. Is there any tricks to tackle this?
 A: You can simplify your expression using the two following identities
ArcTan[a] + ArcTan[b] = ArcTan[(a + b) / (1- a b)]  
Log[(1 - a) / (1 + a)] = 2 ArcTanh[a]
A: A more manageable way should be the following: 
Let us make the change of the variable: $x=\frac{y}{\sqrt{n}}$   
$$a_n=\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx=\frac{2}{n^{\frac{3}{2}}}\int_0^1\frac{y^2}{1+\frac{y^4}{n^2}}dy$$  
Now, let's expand the integrand:  
$$\frac{y^2}{1+\frac{y^4}{n^2}}=\sum_{k=0}^\infty(-1)^k\frac{y^{4k+2}}{n^{2k}}$$ and   
$$\int_0^1\frac{y^2}{1+\frac{y^4}{n^2}}dy=\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\frac{1}{n^{2k}}$$ and  
$$a_n=\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\frac{1}{n^{2k+\frac{3}{2}}}$$
Finally, the sum:  
$$S=\sum_{n=1}^\infty a_n=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\sum_{n=1}^\infty\frac{1}{n^{2k+\frac{3}{2}}}=$$   
$$=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\zeta(2k+\frac{3}{2})$$  
where $\zeta(s)=\sum_{k=1}^\infty\frac{1}{k^s}$ is the Riemann zeta function.
