## Calculus: Early Transcendentals 8th Edition

$$\int x^2\sqrt{2+x}dx=\frac{2}{7}\sqrt{(2+x)^7}-\frac{8}{5}\sqrt{(2+x)^5}+\frac{8}{3}\sqrt{(2+x)^3}+C$$
$$A=\int x^2\sqrt{2+x}dx$$ Sometimes, doing substitution does not eliminate any elements but does simplify the square root so that you can easily transform afterwards. Let $u=2+x$. Then $du=(2+x)'dx=dx$. And since $u=2+x$, we have $x=u-2$, so $x^2=(u-2)^2=u^2-4u+4$. $$A=\int(u^2-4u+4)\sqrt udu$$ $$A=\int(u^2-4u+4)u^{1/2}du$$ $$A=\int(u^{5/2}-4u^{3/2}+4u^{1/2})du$$ $$A=\frac{u^{7/2}}{\frac{7}{2}}-\frac{4u^{5/2}}{\frac{5}{2}}+\frac{4u^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{2\sqrt{u^7}}{7}-\frac{8\sqrt{u^5}}{5}+\frac{8\sqrt{u^3}}{3}+C$$ $$A=\frac{2}{7}\sqrt{(2+x)^7}-\frac{8}{5}\sqrt{(2+x)^5}+\frac{8}{3}\sqrt{(2+x)^3}+C$$