More specifically my integral is
$$\frac{4z\lambda}{4\pi \epsilon} \int_{-a/2}^{a/2} \frac{dx}{(x^2+a^2/4+z^2)^{3/2}}$$
The answer I'm looking for looks something like this: $\frac{1}{4\pi\epsilon}\frac{8\lambda a z}{\sqrt{2a^2+4z^2}(z^2+a^2/4)}$
I really don't see any way to do this.
It is not too hairy. Working temporarily with a single constant $b$ you find an anti-derivative $F(x,b)$ $$\int \frac{dx}{(x^2+b)^{3/2}} = F(x,b)=\frac{x}{b\sqrt{x^2+b}}$$ Now substitute $b$ with your constants $$F\left(x,\frac{a^2}{4}+z^2\right)= \frac{x}{(a^2/4+z^2)\sqrt{x^2+a^2/2+z^2}}$$ $$F\left(\pm\frac{a}{2}, \frac{a^2}{4}+z^2\right) =\pm\frac{1}{2}\frac{a}{(a^2/4+z^2)\sqrt{a^2/2+z^2}} $$ and get for your integral $$ \frac{4z\lambda}{4\pi \epsilon} \int_{-a/2}^{a/2} \frac{dx}{(x^2+a^2/4+z^2)^{3/2}} =\frac{4z\lambda}{4\pi \epsilon}\frac{a}{(a^2/4+z^2)\sqrt{a^2/2+z^2}} =\frac{4z\lambda}{4\pi \epsilon}\frac{2a}{(a^2/4+z^2)\sqrt{2a^2+4z^2}} $$