Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 11

$-\frac 1 {16e^{4x}}(1+4x)+C$

$dv=e^{-4x}dx$ $v=\int e^{-4x}dx = -\frac 1 4e^{-4x}$ $u=x, du=dx$ $\int xe^{-4x}dx = x(-\frac 1 4 e^{-4x})-\int -\frac 1 4e^{-4x}dx$ $\frac x 4 e^{-4x}- \frac 1 {16}e^{-4x}+C$ $-\frac 1 {16e^{4x}}(1+4x)+C$