Chapter A6, Problem 39E

To determine

To verify: the integration formula.

Explanation of Solution

Given:

  ${\displaystyle \int x}{\coth }^{-1}xdx=\frac{x^2-1}{2}{\coth }^{-1}x+\frac{x}{2}+C$

Proof:

To verify, ${\displaystyle \int x}{\coth }^{-1}xdx=\frac{x^2-1}{2}{\coth }^{-1}x+\frac{x}{2}+C$

Take

  $y=\frac{x^2-1}{2}{\coth }^{-1}x+\frac{x}{2}+C$

Differentiate $y$ with respect to $x$ ,

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

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

Hence verified

Conclusion:

Hence the given integration is verified.