## Calculus (3rd Edition)

$$-(4x-3)e^{-x}- 4 e^{-x} +c .$$
We do the integration by parts; we choose $u=4x-3$ and $dv=e^{-x}dx$. Then, $du=4dx$, $v=-e^{-x}$ $$\int (4x-3)e^{-x}dx=\int udv=uv-\int vdu\\ =-(4x-3)e^{-x}+ 4\int e^{-x} dx\\ =-(4x-3)e^{-x}- 4 e^{-x} +c .$$