$\int t e^{a t} d t=\frac{1}{a} t e^{a t}-\frac{1}{a^{2}} e^{a t}+C$

$\int t e^{a t} d t=\frac{1}{a} t e^{a t}-\frac{1}{a^{2}} e^{a t}+C$ , where a is a constant.