Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 512: 10

$\int u^{n}du$ $u=x^{2}-4$ $n=-\frac{1}{2}$

Try it: $u=x^{2}-4$ $du=2xdx$ $\int u^{n}du=\frac{u^{n+1}}{n+1}+C$ $-\frac{u^{n+1}}{n+1}+C$ $-\frac{(x^{2}-4)^{1/2}}{1/2}+C$ $-2(x^{2}-4)^{1/2}+C$ $-2\sqrt {x^{2}-4}+C$ $\int u^{n}du$ $u=x^{2}-4$ $n=-\frac{1}{2}$