<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] This is one case where you can solve the problem without any integration.
Because of the cube in denominator, assume that
$$\int\frac{u^3}{(u^2+1)^3}du=\frac{P_n(u)}{(u^2+1)^2}$$ Differentiate both sides and remove the common denominator to get
$$u^3=\left(u^2+1\right) P_n'(u)-4 u P_n(u)$$ Comparing the degrees $n=2$; so, write $P_2(u)=a+ b u +c u^2$ to get
$$0=b+2 (c-2 a)u-3 b u^2-(2 c+1) u^3$$ Then, $b=0$, $c=-\frac 12$ and $a=-\frac 14$
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