## University Calculus: Early Transcendentals (3rd Edition)

$xe^{x}-e^{x}+C$
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, $\int xe^{x}dx= xe^{x}-\int 1·e^{x}dx= xe^{x}-e^{x}+C$