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 $\pm \infty$?

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
$$ \sin \arctan x = \frac{x}{{\sqrt{x^2+1}}}
$$
