## Calculus: Early Transcendentals 8th Edition

$$\int x^3\sqrt{x^2+1}=\frac{1}{5}\sqrt{(x^2+1)^5}-\frac{1}{3}\sqrt{(x^2+1)^3}+C$$
$$A=\int x^3\sqrt{x^2+1}dx$$ $$A=\int x^2\sqrt{x^2+1}(xdx)$$ Let $u=x^2+1$. Then $du=2xdx$. Therefore, $xdx=\frac{1}{2}du$ And since $u=x^2+1$, we have $x^2=u-1$ Substitute into $A$: $$A=\frac{1}{2}\int(u-1)\sqrt udu$$ $$A=\frac{1}{2}\int(u-1)u^{1/2}du$$ $$A=\frac{1}{2}\int(u^{3/2}-u^{1/2})du$$ $$A=\frac{1}{2}(\frac{u^{5/2}}{\frac{5}{2}}-\frac{u^{3/2}}{\frac{3}{2}})+C$$ $$A=\frac{1}{2}(\frac{2\sqrt{u^5}}{5}-\frac{2\sqrt{u^3}}{3})+C$$ $$A=\frac{\sqrt{u^5}}{5}-\frac{\sqrt{u^3}}{3}+C$$ $$A=\frac{1}{5}\sqrt{(x^2+1)^5}-\frac{1}{3}\sqrt{(x^2+1)^3}+C$$