calculating generalized integral and prove it's convergence So as the title says i want to calculate a generalized integral, but first i want to prove it's convergence, so it has logic behind calculating it.
So the integral is:
$$\int_1^{\infty} \frac{arctanx}{x^4+x^2}dx$$
So i presume it can proven for convergence with bouding arctan between -$\frac{\pi}{2}$ and $\frac{\pi}{2}$
And afer that calcualting, but i'm not sure i'm taking the right route.
Any help solving this would really be appreciated.
 A: For $x\geq 1$,
$$0<\frac{\arctan x}{x^4+x^2}\leq \frac{\pi/2}{2x^2}.$$
Therefore 
$$F(t):=\int_{1}^{t}\frac{\arctan x}{x^4+x^2}\,dx$$
is an increasing function in $[1,+\infty)$. Moreover $F$ is bounded:
$$F(t)\leq \frac{\pi}{4} \int_{1}^{+\infty}\frac{1}{x^2}\,dx=\frac{\pi}{4}\left[-\frac{1}{x}\right]_{1}^{+\infty}=\frac{\pi}{4}.$$
Hence your integral, which is given by $\displaystyle \lim_{t\to+\infty}F(t)$, is convergent. 
To evaluate it, follow the suggestions given by Olivier Oloa. The final result should be
$$\int_{1}^{+\infty}\frac{\arctan x}{x^4+x^2}\,dx=\frac{\ln 2}{2}+\frac{\pi}{4}-\frac{3\pi^2}{32}.$$
A: For the convergence, one may write
$$
I:=\int_1^{\infty} \frac{\arctan x}{x^4+x^2}dx\le\frac \pi2\int_1^{\infty} \frac1{x^4}dx<\infty.
$$
To evaluate $I$, one can make the change of variable $u=\dfrac1x$,
$$
\begin{align}
I=&\int_1^{\infty} \frac{\frac \pi2 -\arctan \frac1x}{1+x^2}\frac{dx}{x^2}
\\\\&=\frac \pi2\int_1^{\infty} \frac{1}{x^2}\frac1{1+\frac{1}{x^2}}\frac{dx}{x^2}-\int_1^{\infty} \frac{1}{x^2}\frac{\arctan \frac1x}{1+\frac{1}{x^2}}\frac{dx}{x^2}
\\\\&=\frac \pi2\int_0^1 \frac{u^2}{1+u^2}du-\int_0^1 \frac{u^2\arctan u}{1+u^2}du
\\\\&=\frac \pi2-\frac \pi2\int_0^1 \frac1{1+u^2}du-\int_0^1 \frac{u^2\arctan u}{1+u^2}du
\end{align}
$$ then conclude, using an integration by parts.
