# How do you evaluate the integral int 1/(xsqrt(1-x^2))?

Jan 30, 2017

The answer is$= - \frac{1}{2} \ln \left(\sqrt{1 - {x}^{2}} + 1\right) + \frac{1}{2} \ln \left(| \sqrt{1 - {x}^{2}} - 1 |\right) + C$

#### Explanation:

We perform this integral by substitution

Let $u = \sqrt{1 - {x}^{2}}$

${u}^{2} = 1 - {x}^{2}$

$\mathrm{du} = \frac{- 2 x \mathrm{dx}}{2 \sqrt{1 - {x}^{2}}} = \frac{- x \mathrm{dx}}{\sqrt{1 - {x}^{2}}}$

Therefore,

$\int \frac{\mathrm{dx}}{x \sqrt{1 - {x}^{2}}} = \int \sqrt{1 - {x}^{2}} \frac{\mathrm{du}}{- {x}^{2} \sqrt{1 - {x}^{2}}}$

$= \int - \frac{\mathrm{du}}{x} ^ 2$

$= \int \frac{\mathrm{du}}{{u}^{2} - 1}$

Now, we perform a de composition into partial fractions

$\frac{1}{{u}^{2} - 1} = \frac{1}{\left(u + 1\right) \left(u - 1\right)} = \frac{A}{u + 1} + \frac{B}{u - 1}$

$= \frac{A \left(u - 1\right) + B \left(u + 1\right)}{\left(u + 1\right) \left(u - 1\right)}$

The denominators are the same, we can compare the numerators

$1 = A \left(u - 1\right) + B \left(u + 1\right)$

Let $u = - 1$, $\implies$, $1 = - 2 A$, $\implies$, $A = - \frac{1}{2}$

Let $u = 1$, $\implies$, $1 = 2 B$, $\implies$, $B = \frac{1}{2}$

Therefore,

$\frac{1}{{u}^{2} - 1} = \frac{- \frac{1}{2}}{u + 1} + \frac{\frac{1}{2}}{u - 1}$

So,

$\int \frac{\mathrm{du}}{{u}^{2} - 1} = - \frac{1}{2} \int \frac{\mathrm{du}}{u + 1} + \frac{1}{2} \int \frac{\mathrm{du}}{u - 1}$

$= - \frac{1}{2} \ln \left(u + 1\right) + \frac{1}{2} \ln \left(u - 1\right)$

$= - \frac{1}{2} \ln \left(\sqrt{1 - {x}^{2}} + 1\right) + \frac{1}{2} \ln \left(| \sqrt{1 - {x}^{2}} - 1 |\right) + C$