## Thomas' Calculus 13th Edition

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, $tanh^{-1} x = tanh^{-1} x$ $(\bf{Verified})$