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.

With $$u=x,v'=e^{-2}$$ we get $$u'=1,v=-\frac{1}{2}e^{-2x}$$ we get
$$\int xe^{2x}dx=-x\frac{1}{2}e^{-2x}-\int\frac{-1}{2}e^{-2x}dx$$
