Question

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

Answer 1

`I = 1/4ln|sqrt(4 - x^2) - 2| - 1/4ln|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 `du = (-2x)/(2sqrt(4 - x^2)) dx` and so `dx = du(sqrt(4 - x^2))/(-x)`

Substituting, we get:

`I = int 1/(xsqrt(4 - x^2)) * sqrt(4 - x^2)/(-x)du`

`I = int -1/x^2du`

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 1/(u^2 - 4) du`

We can now use partial fractions to solve.

`A/(u - 2) + B/(u + 2) = 1/((2 - u)(2 + u))`

`A(2 + u) + B(u - 2) = 1`

`2A + Au -2B + Bu = 1`

`(2A - 2B) + (Au +Bu) = 1`

We can now write a system of equations.

`{(2A - 2B = 1), (A + B = 0):}`

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

`4B = 1`

`B = -1/4`

So `A = 1/4`. The integral now becomes

`I = int 1/(4(u - 2)) - 1/(4(2 + u)) du`

This can be easily integrated now.

`I = 1/4ln|u - 2| - 1/4ln|2 + u| + C`

Reverse our substitutions now.

`I = 1/4ln|sqrt(4 - x^2) - 2| - 1/4ln|2 + sqrt(4 -x^2)| + C`

Hopefully this helps!