Answer
See the proof below.
Work Step by Step
We will have to verify that both sides of the expression are equal:
Thus
$\dfrac{d}{dx} ( \int x coth^{-1} x dx)=\dfrac{d}{dx} ( \dfrac{x^2-1}{2} coth^{-1} x +\dfrac{x}{2} + C)$
or, $ x coth^{-1} x = (\dfrac{x^2 -1}{2}) (\dfrac{1}{1-x^2})+(x)(coth^{-1} x)+\dfrac{1}{2}+0$
Hence, $x coth^{-1} x =x coth^{-1} x $
Therefore, the result has been verified.
