# How do you evaluate the integral int 1/(xsqrt(4-x^2))?

Oct 21, 2017

$I = \frac{1}{4} \ln | \sqrt{4 - {x}^{2}} - 2 | - \frac{1}{4} \ln | 2 + \sqrt{4 - {x}^{2}} | + C$

#### Explanation:

From first glance we would think this integral requires trig substitution but in fact it doesn't. If we let $u = \sqrt{4 - {x}^{2}}$, then $\mathrm{du} = \frac{- 2 x}{2 \sqrt{4 - {x}^{2}}} \mathrm{dx}$ and so $\mathrm{dx} = \mathrm{du} \frac{\sqrt{4 - {x}^{2}}}{- x}$

Substituting, we get:

$I = \int \frac{1}{x \sqrt{4 - {x}^{2}}} \cdot \frac{\sqrt{4 - {x}^{2}}}{- x} \mathrm{du}$

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

But since our bounds of integration are in $u$, this won't quite work. But notice that ${u}^{2} = 4 - {x}^{2}$ and $- {x}^{2} = {u}^{2} - 4$.

$I = \int \frac{1}{{u}^{2} - 4} \mathrm{du}$

We can now use partial fractions to solve.

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

$A \left(2 + u\right) + B \left(u - 2\right) = 1$

$2 A + A u - 2 B + B u = 1$

$\left(2 A - 2 B\right) + \left(A u + B u\right) = 1$

We can now write a system of equations.

$\left\{\begin{matrix}2 A - 2 B = 1 \\ A + B = 0\end{matrix}\right.$

Solving for $A$ and $B$, we get

$4 B = 1$

$B = - \frac{1}{4}$

So $A = \frac{1}{4}$. The integral now becomes

$I = \int \frac{1}{4 \left(u - 2\right)} - \frac{1}{4 \left(2 + u\right)} \mathrm{du}$

This can be easily integrated now.

$I = \frac{1}{4} \ln | u - 2 | - \frac{1}{4} \ln | 2 + u | + C$

Reverse our substitutions now.

$I = \frac{1}{4} \ln | \sqrt{4 - {x}^{2}} - 2 | - \frac{1}{4} \ln | 2 + \sqrt{4 - {x}^{2}} | + C$

Hopefully this helps!