# Evaluate $\int\frac{1}{(\sqrt{d^2 + x^2})^3}dx$

I have to evaluate: $$\int\frac{1}{(\sqrt{d^2 + x^2})^3}dx$$

$x$ is a variable and $d$ is a constant.

I know that $$\int\frac{1}{(d^2 + x^2)}dx$$ has a trivial solution through trigonometric substitution.

However, I'm having some trouble while trying to apply this knowledge to this particular problem. Maybe it has something to do with integration by parts, but I'm clueless about what would be the next step.

Any clever tricks/shortcuts are also welcome.

The solution should be: $$\frac{x}{d^2(\sqrt{d^2 + x^2})} + c$$.

• "I know that "..." has a trivial solution through trigonometric substitution"$$~\\~$$ In that case, have you tried following that same trigonometric substitution $(x=d\tan\theta)$ for this problem? Sep 24, 2017 at 23:17
• Do you know how to solve$$\int\sqrt{d^2+x^2}~\mathrm dx$$? If so, differentiate with respect to $d$ twice. Sep 24, 2017 at 23:35

Factor out the $d$ from the square root and set $\tan\theta = \frac{x}{d}$.
Use the fact that $(1+\tan^2 \theta)$ is $\frac{1}{\cos^2\theta}$, and $dx = d*\frac{\sin \theta}{\cos^2\theta}d\theta$.