Answer
$$\int\left(\sec ^{2} t+ \cot \right)^2 d t=\tan t-2 \ln |\csc t+\cot t|-\cot t-t+C $$
Work Step by Step
Given $$\int\left(\sec ^{2} t+ \cot \right)^2 d t $$
So, we have
We can expand the integral as
\begin{aligned}
I&=\int\left(\sec ^{2} t+ \cot \right)^2 d t\\
&=\int\left(\sec ^{2} t+2 \sec t \cot t+\cot ^{2} t\right) d t\\
&=\int \sec ^{2} t d t+2 \int \sec t \cot t d t+\int \cot ^{2} t d t\\
\end{aligned}
Since, we have
$$
\sec t \cot t=\frac{1}{\cos t} \cdot \frac{\cos t}{\sin t}=\csc t,\\
\cot ^{2} t= \csc ^{2} t-1
$$
So we get
\begin{aligned} I &=\int \sec ^{2} t d t+2 \int \csc t d t+\int\left(\csc ^{2} t-1\right) d t\\
&=\int \sec ^{2} t d t+2 \int \csc t d t+\int \csc ^{2} t d t-\int d t \\
&=\tan t-2 \ln |\csc t+\cot t|-\cot t-t+C
\end{aligned}
