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

In fact, substitution makes it very easy. Starting with
$$\int \frac{u^3}{(u^2 + 1)^3}\ du$$
take $v = u^2 + 1$, then $dv = 2u\ du$ so $u\ du = \frac{1}{2} dv$. Then we get
$$\begin{align}
\int \frac{u^3}{(u^2 + 1)^3}\ du &= \int \frac{u^2 \cdot u}{(u^2 + 1)^3}\ du\\
&= \int \frac{u^2 \cdot \overbrace{(u\ du)}^{\frac{1}{2}\ dv}}{(u^2 + 1)^3}\\
&= \frac{1}{2} \int \frac{u^2\ dv}{v^3}\end{align}$$
Now since $v = u^2 + 1$, we have $u^2 = v - 1$ and
$$\begin{align}
\int \frac{u^3}{(u^2 + 1)^3}\ du &= \frac{1}{2} \int \frac{v - 1}{v^3}\ dv\end{align}$$
which is now easy.
