## Calculus: Early Transcendentals 8th Edition

$$\int x^2\sqrt{x^3+1}dx=\frac{2}{9}\sqrt{(x^3+1)^3}+C$$
$$A=\int x^2\sqrt{x^3+1}dx$$ Let $u=x^3+1$ Then $du=(x^3+1)'dx=3x^2dx$. So $x^2dx=\frac{1}{3}du$. Substitute into $A$, we have $$A=\int \sqrt u(\frac{1}{3})du$$ $$A=\frac{1}{3}\int u^{1/2}du$$ $$A=\frac{1}{3}\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{u^{3/2}}{\frac{9}{2}}+C$$ $$A=\frac{2\sqrt{u^3}}{9}+C$$ $$A=\frac{2}{9}\sqrt{(x^3+1)^3}+C$$