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

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 $z$.  
