How do I solve this integral via integration by parts when it shows a recursion-like behavior? I've come across an integral while solving a question for ODE's. The integral is as follows:
$$\int e^{-2x}x\,dx$$
From here I set:
$$
\begin{align}
f(x) = e^{-2x}\quad & f'(x) = -2e^{-2x} \\
g(x) = \frac{1}{2}x^2\quad & g'(x) = x
\end{align}
$$
and continued to solve the integral as follows:
\begin{align}
\int e^{-2x}x\,dx & = e^{-2x}x - \int \left(-2e^{-2x} \times \frac{1}{2}x^2\right)\,dx \\
& = e^{-2x}x + \int \left( e^{-2x}x^2 \right)\,dx
\end{align}
Solving this integral yields:
$$
\int e^{-2x}x\,dx = e^{-2x}x + \frac{2}{3}\int e^{-2x}x^3\,dx
$$
and it continues like this.
How should I proceed to solve this integral? Any tips or advice is greatly appreciated.

Let $u=x$ and $dv=e^{-2x}dx$. Then, $du=dx$ and $v=\dfrac{e^{-2x}}{-2}$
$$\int e^{-2x}xdx=-\frac{xe^{-2x}}{2}+\frac{1}{2}\int e^{-2x}dx=-\frac{xe^{-2x}}{2}-\frac{e^{-2x}}{4}+C=-\frac{e^{-2x}}{4}\left(2x+1 \right)+C$$
