Integral $\int_0^\infty \frac{t^2\arctan(t)}{(t^2+2)^{{5/2}}}dt$ For integrating $\displaystyle{\int_0^\infty \frac{t^2\arctan(t)}{(t^2+2)^{{5/2}}}dt}\,$ I have tried following substitutions:
i) $t^2+2=(t+x)^{2\over5}$
ii) $(\sqrt2)^2+t^2=(t+x)^{2\over5}$
iii) $t=\sqrt2 \tan(x)$
iv) $t=\sqrt2\sinh(x)$
But every time $\arctan(t)$ term caused problems.What kind of substitution could be applied to this integral?
 A: Integrating by parts might be the best way here. Note that using
$$\int \frac{x}{(x^2+2)^{5/2}}dx=-\frac13 \frac{1}{(x^2+2)^{3/2}}+C$$
We get that the original integral is
$$\int_0^\infty \frac{x^2\arctan x}{(x^2+2)^{5/2}}dx=-\frac13\int_0^\infty x\arctan x\left(\frac{1}{(x^2+2)^{3/2}}\right)'dx$$
$$=0+\frac13 \int_0^\infty \frac{\arctan x}{(x^2+2)^{3/2}}dx+\frac13 \int_0^\infty \frac{x}{(x^2+2)^{3/2}(x^2+1)}dx$$
I'll leave to you the second one since $x^2+2=t^2$ deals with it quite easily. 
For the first one we need to integrate by parts again: 
$$ I_1=\int_0^\infty \frac{\arctan x}{(x^2+2)^{3/2}}dx=\frac12\int_0^\infty \arctan x \left(\frac{x}{(x^2+2)^{1/2}}\right)'dx$$
$$=\frac{\pi}{4}-\frac12\int_0^\infty\frac{x}{(x^2+2)^{1/2}(x^2+1)}dx $$
Now follow up the same substitution as above.
A: Note that
$$ \int\frac{t^2}{(t^2+2)^\frac{5}{2}}dt=\frac16\frac{t^3}{(t^2+2)^\frac32}+C. $$
So by integration by parts, one has
\begin{eqnarray*}
I&=&\int_0^\infty \frac{t^2\arctan(t)}{(t^2+2)^{{5\over2}}}dt\\
&=&\int_0^\infty \arctan(t)d\left(\frac{t^3}{6(t^2+2)^{{3\over2}}}\right)\\
&=&\frac{t^3}{6(t^2+2)^{{3\over2}}}\arctan(t)\bigg|_0^\infty-\frac16\int_0^\infty\frac{t^3}{(t^2+2)^{{3\over2}}(t^2+1)}dt\\
&=&\frac{\pi}{12}-\frac16\int_0^\infty\frac{t^3}{(t^2+2)^{{3\over2}}(t^2+1)}dt.
\end{eqnarray*}
Let $u=t^2+2$ and then
\begin{eqnarray*}
I&=&\frac{\pi}{12}-\frac1{12}\int_{2}^\infty\frac{u-2}{u^{3/2}(u-1)}du
\end{eqnarray*}
which is easy to handle and I omit the detail.
