Chapter A6, Problem 40E

To determine

To verify: the integration formula.

Explanation of Solution

Given:

  ${\displaystyle \int {\tanh }^{-1}}xdx=x{\tanh }^{-1}x+\frac{1}{2}\ln \left(1-x^2\right)+C$

Proof:

To verify, ${\displaystyle \int {\tanh }^{-1}}xdx=x{\tanh }^{-1}x+\frac{1}{2}\ln \left(1-x^2\right)+C$

Take

  $y=x{\tanh }^{-1}x+\frac{1}{2}\ln \left(1-x^2\right)+C$

Now, differentiate $y$ with respect to $x$

  $\Rightarrow \frac{dy}{dx}={\tanh }^{-1}x+x\left(\frac{1}{1-x^2}\right)+\frac{1}{2}\left(\frac{-2x}{1-x^2}\right)$

  $\Rightarrow \frac{dy}{dx}={\tanh }^{-1}x$

Hence verified

Conclusion:

Hence the given integration is verified.