How to integrate: $\frac{1}{(x^2 + z^2)^{\frac{-3}{2}}} dx$ I have tried to use u-substitution but for some reason am not doing it right and thus not getting the correct answer. I want to know the most obvious/ intuitive way to solve this integral.
$x=z\tan(k)$

$\mathrm{d}x=z\sec^{2}(k)\mathrm{d}k$

$(a^2 + x^2)^{\frac32} = z^4 (\sec k)^3 (sec(k))^2 \mathrm{d}k 
=z^4 (1+(\tan k)(\tan k))^{\frac32}(\sec k)^2\mathrm{d}k$ 

now put $\tan(k)=t$, $\mathrm{d}k=(\sec k)(\sec k)\mathrm{d}k$ .
And then integrate it very easily
$x=z$, $\mathrm{d}x=z\sec^{2}(k) \mathrm{d}k$, $(a^2+x^2)^{\frac32}=z^4$