Answer
See the proof below.
Work Step by Step
Verify that both sides of the equation are equal:
$\dfrac{d}{dx} ( \int x tanh^{-1} x dx)=\dfrac{d}{dx} ( x \tanh^{-1} x +\dfrac{1}{2} \ln (1-x^2) + C)$
or, $ tanh^{-1} x = x(\dfrac{1}{1-x^2})+(\tanh^{-1} x) (1)+\dfrac{1}{2}(\dfrac{-2x}{1-x^2})+(0)$
Hence, $tanh^{-1} x = tanh^{-1} x $ $(\bf{Verified})$
