Answer
$$\int \frac{e^{2 x}}{e^{2 x}+3 e^{x}+2} d x =\ln \frac{\left(e^{x}+2\right)^{2}}{e^{x}+1}+C $$
Work Step by Step
Given $$\int \frac{e^{2 x}}{e^{2 x}+3 e^{x}+2} d x$$
\begin{array}{l}{\text { Let } u=e^{x} \text { . Then } x=\ln u, d x=\frac{d u}{u} \Rightarrow} \\ {\qquad \begin{aligned} I&=\int \frac{e^{2 x}}{e^{2 x}+3 e^{x}+2} d x\\
&=\int \frac{u^{2}(d u / u)}{u^{2}+3 u+2}\\
&=\int \frac{u d u}{(u+1)(u+2)}\\
\end{aligned}}\end{array}
Since we have
\begin{array}{l}{ \frac{u }{(u+1)(u+2)} =\frac{A}{(u+1) }+\frac{B}{ (u+2)}
}\\{\text{this implies}}\\{
u=A(u+2)+B(u+1)
}\\{\text{at} \ u=-1 \rightarrow A=-1}\\{\text{at} \ u=-2 \rightarrow B=2}\\{
}\end{array}
So, we get
\begin{aligned}
I&=\int\left[\frac{-1}{u+1}+\frac{2}{u+2}\right] d u \\
&=2 \ln |u+2|-\ln |u+1|+C\\
&= \ln \frac{|u+2|^2}{|u+1|}+C\\
&=\ln \frac{\left(e^{x}+2\right)^{2}}{e^{x}+1}+C
\end{aligned}
