Indefinite integral of rational expression involving cubic polynomials I was given the following exercise:

$$\int \frac{x^3}{(x^2+1)^3}dx$$

As a tip, my professor suggested using the following substitution: $t=x^2+1$. 
Notice that if  $t=x^2+1$, then $x^2=t-1$ and therefore $x=(t-1)^\frac{1}{2}$. Then 
$$x^3=x^2\cdot x=(t-1)(t-1)^\frac{1}{2}=(t-1)^\frac{3}{2}$$
and we have that
$$\int \frac{x^3}{(x^2+1)^3}dx=\int \frac{(t-1)^\frac{3}{2}}{t^3}dt.$$
But now that I have applied the suggested substitution, I don't really see how to continue with this integral. Integrating by parts doesn't seem to get me nowhere, and I can't seem to find any way to apply the substitution method. Am I missing something, or perhaps I made a mistake changing my integration variable from $x$ to $t=x^2+1$?
Thanks in advance.

You forgot that, since $x=(t-1)^{\frac{1}{2}}$, we have also 
$$dx=D((t-1)^{\frac{1}{2}})\, dt=\frac{1}{2}(t-1)^{-\frac{1}{2}}\, dt.$$
Therefore, we find 
$$\int \frac{x^3}{(x^2+1)^3}dx=\int \frac{(t-1)^\frac{3}{2}}{t^3}\cdot
\color{blue}{\frac{1}{2}(t-1)^{-\frac{1}{2}}}\, dt=
\frac{1}{2}\int \frac{t-1}{t^3}\, dt=\frac{1}{2}\int \frac{dt}{t^2}-\frac{1}{2}\int \frac{dt}{t^3}.$$
Now it should be easy to finish the job.
