Answer
$$\int \frac{\cosh t}{\sinh ^{2} t+\sinh ^{4} t} d t=-\frac{1}{\sinh t}-\tan ^{-1}(\sinh t)+C$$
Work Step by Step
Given $$\int \frac{\cosh t}{\sinh ^{2} t+\sinh ^{4} t} d t$$ Let \begin{array}{l}{\mathrm{u}=\text { sinht }} \\ {\mathrm{d} \mathrm{u}=\text { cosht } \mathrm{dt}}\end{array} So, we get \begin{align} I&=\int \frac{1}{u^{4}+u^{2}} d u\\ &=\int \frac{1}{\left(u^{2}\right)\left(u^{2}+1\right)} d u \end{align} Since $$\frac{1}{\left(u^{2}\right)\left(u^{2}+1\right)}=\frac{A u+B}{u^{2}}+\frac{C u+D}{u^{2}+1}$$ So, we get \begin{array}{l}{1=(A u+B)(u^2+1)+(Cu+D}\\{A\left(u^{3}+u\right)+B\left(u^{2}+1\right)+C\left(u^{3}\right)+D\left(u^{2}\right)=1} \\ {A+C=0} \\ {B+D=0} \\ {A=0} \\ {B=1} \\ {C=0} \\ {D=-1}\end{array} This implies \begin{align}I&=\int \frac{1}{\left(u^{2}\right)\left(u^{2}+1\right)} d u\\ &=\int\left(\frac{1}{u^{2}}-\frac{1}{u^{2}+1}\right) d u\\ &=-\frac{1}{u}-\tan ^{-1} u+C\\ &=-\frac{1}{\sinh t}-\tan ^{-1}(\sinh t)+C\end{align}
