Calculus (3rd Edition)

$\frac{11}{16}e^{12}+\frac{1}{16}$
We do integration by parts; let $u=x$ and $dv=e^{4x}dx$ then $du =dx$ and $v=\frac{1}{4}e^{4x}$. Now, we get $$\int_{0}^{3} x e^{4 x} d x= \int_{0}^{3} udv=uv|_{0}^{3}-\int_{0}^{3} vdu\\ =\frac{1}{4}xe^{4x}|_{0}^{3}-\frac{1}{16} e^{4x}|_{0}^{3}\\ =\frac{3}{4}e^{12}-\frac{1}{16}e^{12}+\frac{1}{16} =\frac{11}{16}e^{12}+\frac{1}{16}.$$