## University Calculus: Early Transcendentals (3rd Edition)

We need to verify both sides of the expression. In order to do this, we will differentiate both sides. $\dfrac{d}{dx} ( \int x tanh^{-1} x dx)=\dfrac{d}{dx} ( x \tanh^{-1} x +\dfrac{1}{2} \ln (1-x^2) + C)$ or, $tanh^{-1} x = x(\dfrac{1}{1-x^2})+(\tanh^{-1} x) (1)+\dfrac{1}{2}(\dfrac{-2x}{1-x^2})+(0)$ or, $tanh^{-1} x = tanh^{-1} x$ Hence, the result has been verified.