<s>[INST] How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution?  Thank you
 [/INST] Letting
$v = u^2+1$,
$dv = 2u\, du, u^2 = v-1$
so
$\begin{array}\\
\int\dfrac{u^3}{(u^2+1)^3}du
&=\int\dfrac{(v-1)}{2v^3}dv\\
&=\frac12\int(v^{-2}-v^{-3}dv\\
&=\frac12\left(\dfrac{v^{-1}}{-1}-\dfrac{v^{-2}}{-2}\right)\\
&=\frac12\left(-v^{-1}+\frac12 v^{-2}\right)\\
&=\frac12\left(-\dfrac1{u^2+1}+\dfrac1{2(u^2+1)^2}\right)\\
&=\frac12\dfrac{-2(u^2+1)+1}{2(u^2+1)^2}\\
&=\dfrac{-2u^2-1}{4(u^2+1)^2}\\
&=-\dfrac{2u^2+1}{4(u^2+1)^2}\\
\end{array}
$
</s>