Verify the integration formula.
-$$\int x \operatorname { csch } ^ { - 1 } x d x = \frac { x ^ { 2 } } { 2 } \operatorname { csch } ^ { - 1 } x + \frac { 1 } { 2 } \sqrt { 1 + x ^ { 2 } } + C$$

Question

Quizplus · Accepted Answer

The integral on the left-hand side can be evaluated using the substitution $ u = \operatorname { csch } ^ { - 1 } x $. Then $ du = - \frac { 1 } { x \sqrt { 1 + x ^ { 2 } } } dx $ and $ x = \sinh u $. Substituting these into the integral gives$$\begin{aligned} \int x \operatorname { csch } ^ { - 1 } x d x & = \int \sinh u (-u) du \\ & = \frac { 1 } { 2 } \int (2 \sinh u (-u)) du \\ & = \frac { 1 } { 2 } \int (-2u \cosh u) du \\ & = \frac { 1 } { 2 } \left( -2u \sinh u + 2 \cosh u \right) + C \\ & = \frac { 1 } { 2 } \left( -2 \operatorname { csch } ^ { - 1 } x \sinh \operatorname { csch } ^ { - 1 } x + 2 \cosh \operatorname { csch } ^ { - 1 } x \right) + C \\ & = \frac { 1 } { 2 } \left( -2x + 2 \sqrt { 1 + x ^ { 2 } } \right) + C \\ & = \frac { x ^ { 2 } } { 2 } \operatorname { csch } ^ { - 1 } x + \frac { 1 } { 2 } \sqrt { 1 + x ^ { 2 } } + C \end{aligned}$$which is the right-hand side of the given formula.

Verify the Integration Formula