Calculus: Repeated Integration by Parts

Using $\int u v' \, dx = uv - \int u' v \, dx$. Set $u = x^2 \implies u' = 2x$, and $v' = e^{2x} \implies v = \frac{1}{2}e^{2x}$: $$\int x^2 e^{2x} \, dx = x^2\left(\frac{1}{2}e^{2x}\right) - \int (2x)\left(\frac{1}{2}e^{2x}\right) \, dx = \frac{1}{2}x^2 e^{2x} - \int x e^{2x} \, dx$$

Set $u = x \implies u' = 1$, and $v' = e^{2x} \implies v = \frac{1}{2}e^{2x}$: $$\int x e^{2x} \, dx = x\left(\frac{1}{2}e^{2x}\right) - \int (1)\left(\frac{1}{2}e^{2x}\right) \, dx = \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x}$$

$$\int x^2 e^{2x} \, dx = \frac{1}{2}x^2 e^{2x} - \left( \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} \right) + C$$ $$= e^{2x}\left( \frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{4} \right) + C = \frac{1}{4}e^{2x}(2x^2 - 2x + 1) + C$$