## Calculus, 10th Edition (Anton)

$\dfrac{2}{3}(\sqrt{10}-\sqrt{8})$
In order to integrate this function, we have to use an «u» substitution. Choose an u, derivate it: $u =x^3+9$ $du=3x^2dx$ $\dfrac{1}{3}du =x^2dx$ Change limits: $x_1 = -1$ $u_1= (-1)^3+9=8$ $x_2 = 1$ $u_2=(1)^3+9=10$ Then substitute: $\dfrac{1}{3} \int_{8}^{10} u^{-1/2} du$ Then integrate $\dfrac{1}{3} \bigg [ \dfrac{u^{1/2}}{1/2} \bigg ]_{8}^{10}$ Apply calculus 2nd theorem and reduce: $\dfrac{2}{3} (\sqrt{10}-\sqrt{8})$