Answer
$$\int \frac{\sec ^{2} t}{\tan ^{2} t+3 \tan t+2} d t=\ln \frac{ |\tan t+1|}{ |\tan t+2|}+C$$
Work Step by Step
Given $$\int \frac{\sec ^{2} t}{\tan ^{2} t+3 \tan t+2} d t$$
Let
\begin{array}{l}{ u=\tan t, \ \ \ d u=\sec ^{2} t \ d t . \text { Then }\\
\begin{aligned}I&=\int \frac{\sec ^{2} t}{\tan ^{2} t+3 \tan t+2} d t\\
&=\int \frac{1}{u^{2}+3 u+2} d u\\
&=\int \frac{1}{(u+1)(u+2)} d u\end{aligned}} \\ {\text { Now: }\\ \frac{1}{(u+1)(u+2)}=\frac{A}{u+1}+\frac{B}{u+2} \Rightarrow 1=A(u+2)+B(u+1)} \\ {\text { Setting } u=-2\Rightarrow 1=-B, \Rightarrow B=-1}\\{ \text { Setting } u=-1\Rightarrow A=1} \\
{\text { Thus, } \\
\begin{aligned} I&=\int \frac{1}{(u+1)(u+2)} d u\\
&=\int\left(\frac{1}{u+1}-\frac{1}{u+2}\right) d u\\
&=\ln |u+1|-\ln |u+2|+C\\
&=\ln |\tan t+1|-\ln |\tan t+2|+C \\
&=\ln \frac{ |\tan t+1|}{ |\tan t+2|}+C\end{aligned}}\end{array}