Интеграл (x+1)sin(x^2+2x)

Решите задачу:

$\displaystyle \int \left(x+1\right)\sin \left(x^2+2x\right)dx\\ \\ u=x^2+2x\\ \\ \int \frac{\sin \left(u\right)}{2}du=\frac{1}{2}\cdot \int \:\sin \left(u\right)du=\frac{1}{2}\left(-\cos \left(u\right)\right)=\frac{1}{2}\left(-\cos \left(x^2+2x\right)\right)=\\ \\ =-\frac{1}{2}\cos \left(x^2+2x\right)=-\frac{1}{2}\cos \left(x^2+2x\right)+C$