Answer
$$t+\frac{1}{4} \operatorname{coth}(1-4 t)+C$$
Work Step by Step
We integrate as follows:
\begin{aligned}
\int \operatorname{coth}^{2}(1-4 t) d t &=\int\left(1+\operatorname{csch}^{2}(1-4 t)\right) d t \\
&=\int 1 \, d t+\int \operatorname{csch}^{2}(1-4 t) d t \\
&=t+\frac{1}{4} \operatorname{coth}(1-4 t)+C
\end{aligned}
