Answer
$\int(x+1)e^{-(x^2+2x)}dx=-\frac{1}{2}e^{-(x^2+2x)}+C$
Work Step by Step
Substitution:
$u=-(x^2+2x)$
$\frac{du}{dx}=-(2x+2)=-2(x+1)$
$dx=-\frac{1}{2(x+1)}du$
$\int(x+1)e^{-(x^2+2x)}dx=\int(x+1)e^u(-\frac{1}{2(x+1)}du)=\int-\frac{1}{2}e^udu=-\frac{1}{2}e^u+C=-\frac{1}{2}e^{-(x^2+2x)}+C$
