Calculus: Repeated Integration by Parts

Let $u = x^2 \implies du = 2x \, dx$. Let $dv = e^{3x} \, dx \implies v = \frac{1}{3}e^{3x}$. $$\int x^2 e^{3x} \, dx = \frac{1}{3}x^2 e^{3x} - \int \frac{1}{3}e^{3x}(2x) \, dx = \frac{1}{3}x^2 e^{3x} - \frac{2}{3} \int x e^{3x} \, dx$$

For $\int x e^{3x} \, dx$, let $u_1 = x \implies du_1 = dx$, $dv_1 = e^{3x} \, dx \implies v_1 = \frac{1}{3}e^{3x}$. $$\int x e^{3x} \, dx = \frac{1}{3}x e^{3x} - \int \frac{1}{3}e^{3x} \, dx = \frac{1}{3}x e^{3x} - \frac{1}{9}e^{3x}$$

$$\int x^2 e^{3x} \, dx = \frac{1}{3}x^2 e^{3x} - \frac{2}{3}\left( \frac{1}{3}x e^{3x} - \frac{1}{9}e^{3x} \right) + C$$ $$= \frac{1}{3}x^2 e^{3x} - \frac{2}{9}x e^{3x} + \frac{2}{27}e^{3x} + C = \frac{e^{3x}}{27}(9x^2 - 6x + 2) + C$$