Answer
$\dfrac{x e^{2x}}{2}-\dfrac{1}{4} e^{2x} -\dfrac{4e^{3x}}{3}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int (x e^{2x}-4e^{3x}) \ dx=\dfrac{x e^{2x}}{2}-\dfrac{1}{2} \int e^{2x} \ dx-\int 4e^{3x} \ dx$
or, $=\dfrac{x e^{2x}}{2}-\dfrac{1}{4} -\int 4e^{3x} \ dx+C$
or, $=\dfrac{x e^{2x}}{2}-\dfrac{1}{4} e^{2x} -4 \times (\dfrac{e^{3x}}{3})+C$
or, $=\dfrac{x e^{2x}}{2}-\dfrac{1}{4} e^{2x} -\dfrac{4e^{3x}}{3}+C$
