Question

Integration xe^(-0.15x)

Answer 1

$\int\limits {xe^{-0.15x}} \, dx$
Let$-0.15 x=t$
∴$-0.15 dx = dt$ and $x = \frac{t}{-0.15}$

$\int\limits { \frac{t}{-0.15}*e^t* \frac{1}{-0.15} } \, dt$
$\frac{1}{0.0225} \int\limits {te^t} \, dt$
Using Integration by parts we get...
$\frac{1}{0.0225}*(te^t-e^t) + C$
$\frac{1}{0.0225}*e^{-0.15x}*(-0.15x-1) + C$