This integral: $$\int \frac{dx}{(z^2 + x^2)^\frac{3}{2}}$$ comes up a lot. My only method of solving this using the substitution $x=z\tan(\theta)$. It works fine but I was wondering if there was any other clever ways of solving this integral or any very similar integral?
$$\int \frac{dx}{(z^2 + x^2)^\frac{3}{2}}=\frac{x}{z^2\sqrt{(z^2 + x^2)}} + x_0$$
On
I'm using the OP's original variable of integration $z$.
$$z=xt$$
$$\int \frac{dz}{(z^2 + x^2)^\frac{3}{2}}=\frac{1}{x^2} \int \frac{dt}{(1 + t^2)^\frac{3}{2}}$$
$$1+t^2=s^2, \qquad t=\sqrt{s^2-1}, \qquad dt=\frac{sds}{\sqrt{s^2-1}}$$
$$\int \frac{dt}{(1 + t^2)^\frac{3}{2}}=\int \frac{ds}{s^2\sqrt{s^2-1}}$$
$$s=\frac{1}{y}, \qquad ds=-\frac{dy}{y^2}$$
$$\int \frac{ds}{s^2\sqrt{s^2-1}}=-\int \frac{ydy}{\sqrt{1-y^2}}$$
I think it's absolutely obvious that:
$$-\int \frac{ydy}{\sqrt{1-y^2}}=\sqrt{1-y^2}$$
Now getting back to our original variables:
$$y=\frac{1}{s}=\frac{1}{\sqrt{1+t^2}}=\frac{x}{\sqrt{x^2+z^2}}$$
$$1-y^2=1-\frac{x^2}{x^2+z^2}=\frac{z^2}{x^2+z^2}$$
$$\sqrt{1-y^2}=\frac{z}{\sqrt{x^2+z^2}}$$
Finally:
$$\int \frac{dz}{(z^2 + x^2)^\frac{3}{2}}=\frac{1}{x^2} \frac{z}{\sqrt{x^2+z^2}}$$
On
Here is another way.
Let $$x = \frac{1}{u}, \quad dx= -\frac{1}{u^2} \, du.$$ So the integral becomes $$\int \frac{dx}{(z^2 + x^2)^{3/2}} = - \int \frac{u}{(z^2 u^2 + 1)^{3/2}} \, du.$$
Now let $$t = z^2 u^2 + 1, \quad u \, du = \frac{dt}{2z^2}.$$ Thus $$\int \frac{dx}{(z^2 + x^2)^{3/2}} = -\frac{1}{2z^2} \int t^{-3/2} \, dt = \frac{1}{z^2 \sqrt{t}} + C = \frac{1}{z^2 \sqrt{z^2 u^2 + 1}} + C,$$ or $$\int \frac{dx}{(z^2 + x^2)^{3/2}} = \frac{x}{z^2 \sqrt{z^2 + x^2}} + C.$$
Hint...try $z=x\sinh\theta$ instead