Answer
$$-\frac{12 x+1}{144} e^{-12 x}+C$$
Work Step by Step
Given $$ \int x e^{-12 x} d x$$
Let
\begin{align*}
u&=x\ \ \ \ \ \ \ dv =e^{-12x}dx\\
du&=dx\ \ \ \ \ v=\frac{-1}{12}e^{-12x}
\end{align*}
Then
\begin{aligned} \int x e^{-12 x} d x &=x \cdot\left(-\frac{e^{-12 x}}{12}\right)-\int\left(-\frac{e^{-12 x}}{12}\right) \cdot 1 d x \\ &=-\frac{1}{12} x e^{-12 x}+\frac{1}{12} \int e^{-12 x} d x \\ &=-\frac{1}{12} x e^{-12 x}+\frac{1}{12} \cdot\left(-\frac{e^{-12 x}}{12}\right)+C \\ &=-\frac{1}{12} x e^{-12 x}-\frac{1}{144} e^{-12 x}+C \\ &=-\frac{12 x+1}{144} e^{-12 x}+C \end{aligned}
