Answer
$\dfrac {\left( 1+e^{x}\right) ^{3}}{3}+C$
Work Step by Step
$u=1+e^{x}\Rightarrow du=e^{x}dx$
$$\int \left( 1+e^{x}\right) ^{2}e^{x}dx=\int u^{2}du=\dfrac {u^{3}}{3}+C=\dfrac {\left( 1+e^{x}\right) ^{3}}{3}+C$$
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