Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.6 Exercises - Page 555: 19

$\int\frac{1}{x^{2}\sqrt{x^{2}-4}}dx $ $=$ $-$ $\frac{\sqrt{x^{2}-4}}{4x}$ $+$ $c $

$\int\frac{1}{x^{2}\sqrt{x^{2}-4}}dx $ $Let $ $u=x $ $, $ $du=dx $ $with $ $a=$ $, $ $a^{2}=4$ $so $ $\int\frac{1}{x^{2}\sqrt{x^{2}-4}}dx $$=$$\int\frac{du}{u^2\sqrt{u^2-a^2}}$ $=-$ $\frac{\sqrt{u^2-a^2}}{a^2 u}$ $+c $ $By $ $substituion $ $we $ $have $ $:$ $\int\frac{1}{x^{2}\sqrt{x^{2}-4}}dx $ $=$ $-$ $\frac{\sqrt{x^{2}-4}}{4x}$ $+$ $c $