Answer
$\dfrac{2\pi}{3}$
Work Step by Step
$\iiint_{x^2+y^2 \leq 1} 4x(x+2) (x^2+y^2) dy dx=\int_{0}^{2 \pi}\int_0^{1} 4(r \cos \theta) (r \cos \theta+2)r^2 dr d\theta$
$=(4) \times \int_{0}^{2 \pi}\int_0^{1} (r^2 \cos^2 \theta+2r \cos \theta) r^3 dr d\theta$
$=(4) \times \int_{0}^{2 \pi}\int_0^{1} r^5 \cos^2 \theta dr d\theta+8 \times \int_{0}^{2 \pi}\int_0^{1} r^4 \cos \theta dr d\theta$
$=4[\dfrac{\theta}{2}+\dfrac{\sin 2\theta}{4}]_0^(2 \pi) \times (r^6/6)_{0}^{1}+8 \times [\sin \theta]_0^{2 \pi} (r^5/5)_0^1$
$=\dfrac{2\pi}{3}$
