Answer
$$\int(2\theta+1+2\cos(2\theta+1))d\theta=\frac{(2\theta+1)^2}{4}+\sin(2\theta+1)+C$$
Work Step by Step
$$A=\int(2\theta+1+2\cos(2\theta+1))d\theta$$
We set $a=2\theta+1$, then $$da=2d\theta$$ $$d\theta=\frac{1}{2}da$$
Therefore, $$A=\frac{1}{2}\int (a+2\cos a)da$$ $$A=\frac{1}{2}\Big(\int ada+2\int\cos ada\Big)$$ $$A=\frac{1}{2}\Big(\frac{a^2}{2}+2\sin a\Big)+C$$ $$A=\frac{a^2}{4}+\sin a+C$$ $$A=\frac{(2\theta+1)^2}{4}+\sin(2\theta+1)+C$$
