What's a method for computing the indefinite integral $\int \dfrac{dz}{(a^2 + z^2)^{3/2}}$?

Question

This integral occurs in EMFT when computing $\overline{E}$ due to an infinite line, uniform charge distribution.

I'm trying to figure out the formula for $\int\dfrac{dz}{(a^2 + z^2)^{3/2}}$, using integration by parts but fail.

Choosing $dv = (a^2 + z^2)^{-3/2}dz$ leaves us within the same boat we started.

Choosing $u = (a^2 + z^2)^{-3/2}, \ dv = dz$, gives: $$ du = -(3/2)(a^2 + z^2)^{-5/2}dx $$ so that $$ \int u dv = uv - \int v du = z(a^2 + z^2)^{-3/2} - \int \dfrac{2z^2}{(a^2 + z^2)^{5/2}}dz $$ Leaving me with a mess.

What's the trick to this one? Does it help to know that the limits of integration are $\pm \infty$?

Answer 1

Hint:

Perform the trigonometric substitution $z=a\tan u$.

Edit:

When $z=a\tan u$, $dz=a\sec^2 u du$

$$\displaystyle \int \dfrac{dz}{(a^2 + z^2)^{3/2}}=\int \dfrac{a\sec^2 u du}{(a^2 + a^2\tan^2u)^{3/2}}=\int \dfrac{a\sec^2 u du}{(a^2(1+\tan^2u))^{3/2}}=\int \dfrac{a\sec^2 u du}{a^3\sec^3u}=\frac{1}{a^2}\int \dfrac{du}{\sec u}=\frac{1}{a^2}\int \cos udu=\frac{\sin u}{a^2}+C$$

Now, back substitute for $z$.

Answer 2

Let $a = 1$ for convenience. You can generalise it to arbitrary $a$ by the method of substitution.

Let $z = \tan w,dz = (\sec w)^2 dw$. Then $$ \int \frac{dz}{(z^2 + 1)^{3/2}} = \int \frac{(\sec w)^2}{((\tan w)^2 + 1)^{3/2}} dw $$ Since $(\tan w)^2 + 1 = (\sec w)^2$, $$ = \int \frac{dw}{\sec w} = \int \cos w\,dw = \sin w + C $$ and simplify using the identity $$ \sin \arctan x = \frac{x}{{\sqrt{x^2+1}}} $$