I would definitely recommend $w=1+2x$ to begin. Linear substitutions like that cost you nothing, and could gain you quite a bit, as in this case. What is the resulting $w$ integral?
Another way to solve this integral is to notice that
\begin{align*}
\frac{d}{dx} \frac{f(x)}{g(x)}&=\frac{g(x)\, f'(x)-f(x) \, g'(x)}{(g(x))^{2}} \\
\frac{f(x)}{g(x)}+C&=\int \frac{g(x)\, f'(x)-f(x) \, g'(x)}{(g(x))^{2}} \, dx.
\end{align*}
The quotient rule doesn't come in handy all that often, but when it does, it surprises you.