Question
Evaluate: $\int \frac{e^{2 x}}{2+e^x} d x$
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Answer
Given, $\int \frac{e^{2 x}}{2+e^x} d x$
Let, $e^x+2=t$
$e^x=t-2$
$\Rightarrow \quad e^x d x=d t$
$\therefore \quad \int \frac{t-2}{t} d t=t-2 \log |t|+C_1$
$\begin{array}{l}=e^x+2-2 \log |t|+C_1 \\ =e^x-2 \log \left|e^x+2\right|+C\end{array}$
$\left[\right.$ Since $\left.e^{2 x} d x=e^x \cdot e^x d x=(t-2) d t\right]\left( C = C _1+2\right)$
Let, $e^x+2=t$
$e^x=t-2$
$\Rightarrow \quad e^x d x=d t$
$\therefore \quad \int \frac{t-2}{t} d t=t-2 \log |t|+C_1$
$\begin{array}{l}=e^x+2-2 \log |t|+C_1 \\ =e^x-2 \log \left|e^x+2\right|+C\end{array}$
$\left[\right.$ Since $\left.e^{2 x} d x=e^x \cdot e^x d x=(t-2) d t\right]\left( C = C _1+2\right)$
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