Indefinite integral of rational expression involving cubic polynomials

Question

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.

Answer 1

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.

Answer 2

Partial integration may also work as follows:

\begin{eqnarray*} \int \frac{x^3}{(x^2+1)^3}dx & = & \int x^2\frac{x}{(x^2+1)^3}dx \\ & = & x^2\left(-\frac{1}{4}\frac{1}{(x^2+1)^2}\right) + \frac{1}{4}\int \frac{2x}{(x^2+1)^2}dx \\ & = & -\frac{1}{4}\frac{x^2}{(x^2+1)^2} - \frac{1}{4}\frac{1}{x^2+1} (+C) \end{eqnarray*}