$$\int \frac{3}{2} x e^{-6 x} d x$$
Hi, I was solving an equation and I came to a part requiring that I solve $\int{\frac{3}{2}xe^{-6x}dx}$. I attempted to solve this using $u$-sub but got the wrong answer, I set $u = e^{-6x}$.
$$ \quad \begin{aligned} \text { Let } u & =e^{-6 x} d u \\ d u & =-6 e^{-6 x} d x \\ \frac{d u}{6} & =e^{-6 x} d x \end{aligned} \\ \int \frac{3}{2} x e^{-6 x} dx\to\frac{3}{2} \int x \frac{d u}{6} \\ \to\frac{3}{12} \int{x}=\frac{3}{12} \cdot \frac{x^2}{2}=\frac{3 x^2}{24}=\frac{1}{8} x^2 \\ $$
I was wondering why this strategy didn't work. Is it because after the $u$-sub the integral is with respect to $u$ and I was left with an $x$?
Note: If the integrand is a logarithm, inverse trigonometric function, inverse hyperbolic function, or a product of $x^m$ with $\ln x$, $e^{ax}$, $\sin{ax}$ or $\cos{ax}$ or their inverses, then the solution can be got by a single or repeated integration by parts.
Hence, in your problem we can choose: $u=x \implies du=dx,\, $ and $\, dv=e^{-6x}dx \implies v=-\frac{1}{6}e^{-6x}$.
Therefore we have: $$\int \frac{3}{2}xe^{-6x}\mathrm dx=\frac{3}{2}\left(-\frac{1}{6}xe^{-6x}+\frac{1}{6}\int e^{-6x}\mathrm dx \right)=-\frac{1}{4}xe^{-6x}-\frac{1}{24}e^{-6x}=\boxed{-\frac{1}{4}e^{-6x}\left(x+\frac{1}{6} \right)}.$$