Calculus: Early Transcendentals 8th Edition

$\frac{-xe^{2x}}{2(1+2x)}+\frac{1}{4}e^{2x}+C$
Choose $u=xe^{2x}$ and $dv={(1+2x)}^{-2} \ dx$. To solve for $du$, use the product rule. $$\frac{d}{dx}xe^{2x}=2xe^x+e^{2x}=(2x+1)e^{2x}$$ Solving for $v$: $$\int(1+2x)^{-2} \ dx=\frac{-1}{2(1+2x)}$$ Now apply the integration by parts formula ($uv-\int v \ du$) $$\frac{-xe^{2x}}{2(1+2x)}-\int \frac{-1}{2(1+2x)} \cdot (2x+1)e^{2x} \ dx$$ $$=\frac{-xe^{2x}}{2(1+2x)}+ \frac{1}{2} \int e^{2x} \ dx$$ $$=\frac{-xe^{2x}}{2(1+2x)} + \frac{1}{4} e^{2x} +C$$