Answers

2016-03-11T10:09:05+05:30
 \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
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