How do we solve the given integral to reach the given solution?

Question

I encountered this integral given below.

$$\int_{-1}^1 \frac{z-Ru}{(R^2+z^2-2Rzu)^{3/2}}\mathop {\mathrm du}$$

The recommended answer is $$\frac1{z^2}\left\{\frac{z-R}{|z-R|}-\frac{-z-R}{|z+R|}\right\}$$

Although I can solve this by separating the terms in the numerator and then substituting for the expression in the denominator within the brackets as $y^2$. This will pose two conditions: either $z>R$ or $R>z$, in which case I must find the final expression for each condition separately.

However, I am desperate to know how to obtain the elegant expression as given in the recommended answer.

Answer 1

First consider the integral $$\int_{-1}^1 \frac{du}{\sqrt{R^2+z^2+2Rzu}}$$ This one is rather easy to calculate an antiderivative for; it is equal to $$\int \frac{du}{\sqrt{R^2+z^2+2Rzu}}=\frac{1}{Rz}\sqrt{R^2+z^2+2Rzu}$$ and so evaluating across the given limits of integration gives us the answer $$\int_{-1}^1 \frac{du}{\sqrt{R^2+z^2+2Rzu}}=\frac{\sqrt{(R+z)^2}-\sqrt{(R-z)^2}}{Rz}$$ Differentiate both sides of this with respect to $z$ and the answer should follow, because the LHS will become $$-\int_{-1}^1 \frac{z+Ru}{\sqrt{R^2+z^2+2Rzu}}du$$ which turns in to the integral you are looking for if you substitute $z\to -z$.