Answer
$$\int \frac{coth(1/y)}{y^{2}}dy=-ln|sinh\,\frac{1}{y}|+C$$
Work Step by Step
Table of integrals #106:
$\int coth\,u\,du=ln|sinh\,u|+C$
$let\,u=1/y,\,du=-\frac{dy}{y^{2}}$
$$\int \frac{coth(1/y)}{y^{2}}dy=-\int coth\,u\,du$$
$$=-ln|sinh\,u|+C$$
$$=-ln|sinh\,\frac{1}{y}|+C$$
