Answer
$\frac{-\sqrt {1-x^2}}{x} +C$
Work Step by Step
Use Integration by tables to find the indefinite integral
$\int \frac{1}{x^2\sqrt {1-x^2}}dx$
Let $ a=1, u=x, du=dx$
Rewrite the integrand
$\int \frac{1}{u^2\sqrt {a^2-u^2}}du$
Use Formula #44 in the integration tables to integrate
$\int \frac{1}{u^2\sqrt {a^2-u^2}}du= \frac{-\sqrt {a^2-u^2}}{a^2u}+C$
Resubstitute
$\frac{-\sqrt {1-x^2}}{x} +C$
