<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] Substitute $v=u^2+1$ so that $du=\frac{1}{2u}dv$ to turn the integral into:
$$\int{\frac{u(v-1)}{2uv^3}} dv$$
$$\frac{1}{2}\int{\frac{v-1}{v^3}} dv$$
$$=\frac{1}{2}(\int{\frac{1}{v^2} dv}-\int{\frac{1}{v^3}} dv)$$
$$=\frac{1}{2}(\frac{1}{2v^2}-\frac{1}{v})+C$$
$$=\frac{1}{4v^2}-\frac{1}{2v}+C$$
Substitute $v=u^2+1$ and simplify to get your answer:
$$\dfrac{1}{4(u^2+1)^2}-\dfrac{1}{2(u^2+1)}+c$$
$$=-\dfrac{2u^2+1}{4\left(u^2+1\right)^2}+C$$
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