Answer
$\dfrac{1}{2} \sin (x^2+2x)+\dfrac{ 1}{4} (x^2+2x)^2+C$
Work Step by Step
Our aim is to solve the integral $\int (x+1 ) [\cos (x^2+2x)+(x^2+2x)] \ dx$
Let us consider that $a =x^2+2x$ and $\dfrac{da}{dx}=2x+2 \implies dx=\dfrac{1}{2(x+1)} \ da$
Now, $\int (x+1 ) [\cos (x^2+2x)+(x^2+2x)] \ dx = \int (x+1 ) (\cos a+a)\times (\dfrac{1}{2(x+1)} ) \ da$
or, $=\dfrac{1}{2}\int (\cos a +a) \ da$
or, $=\dfrac{1}{2} \sin (x^2+2x)+\dfrac{ 1}{4} (x^2+2x)^2+C$
