definite integral $\int_0^\infty\frac{r^3}{(r^2+d^2)^3} dr$

Question

$$\int_0^\infty\frac{r^3}{(r^2+d^2)^3} dr$$

I am completely stumped how to solve this integral for a physics problem. I thought about trying to write it in terms of partial fractions with denominators that are multiples of $(r-d)$ and $(r+d)$ but I got lost in the algebra.

Are there any suitable substitutions that might help solve this integral?

Answer 1

$$\begin{align} \\ & \int_0^\infty\frac{r^3}{(r^2+d^2)^3} dr \\ & =\frac{1}{2}\cdot \int_0^\infty\frac{r^2}{(r^2+d^2)^3}\cdot (2rdr) \end{align}$$

Now we take $z=r^2+d^2$ and hence $dz=2rdr$.

Also when $r=o$, $z=d^2$ and when $r\to \infty$, $z\to \infty$.

Therefore we have that

$$\begin{align} \\ & = \frac{1}{2}\cdot \int_{d^2}^\infty \frac{z-d^2}{z^3}dz \\ & = \frac{1}{2}\cdot \int_{d^2}^\infty\frac{z-d^2}{z^3}dz \\ & = \frac{1}{2}\cdot \int_{d^2}^\infty \frac{1}{z^2}dz - \frac{1}{2}\cdot \int_{d^2}^\infty \frac{d^2}{z^3}dz \\ & = \frac{1}{2}\cdot \frac{1}{d^2} - \frac{d^2}{2}\cdot \frac{1}{2d^4}dz \\ & = \frac{1}{2d^2} - \frac{1}{4d^2} \\ & = \frac{1}{4d^2} \end{align}$$

Hope this helps.

Answer 2

Write it as:

$$r^2 \times \frac{r}{(r^2 + d^2)^3}$$

Then integrate by parts.