Answer
$xe^{x}-e^{x}+C$
Work Step by Step
We use the formula:
$\int f(x)g(x)dx=$
$f(x)\int g(x)dx-\int [f'(x)\int g(x)dx]dx$.
Put f(x)=x and g(x)=$ e^{x}$.
The derivative of first function is 1 and the integral of the second function is $e^{x}$.
Therefore,
$\int xe^{x}dx= xe^{x}-\int 1·e^{x}dx= xe^{x}-e^{x}+C$
