Question

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

Answer 1

`1/4lnabs((x-2)/(x+2))+C`

Explanation:

This can be done with partial fractions, but this will be done with a trigonometric substitution.
First, we need to add `dx` to the integral:

`I=int1/(x^2-4)dx`

Recall that `sec^2theta-1=tan^2theta`.

Because of that identity, let `x=2sectheta`.

Importantly, this implies that `x^2-4=4sec^2theta-4=4(sec^2theta-1)=4tan^2theta`.

Furthermore, `dx=2secthetatanthetad theta`.

Then:

`I=int(2secthetatantheta)/(4tan^2theta)d theta=1/2intsectheta/tanthetad theta=1/2intcscthetad theta`

Which is a commonly known integral:

`I=-1/2lnabs(csctheta+cottheta)`

From our original substitution `x=2sectheta` we see that `sectheta=x/2`. This is a right triangle where `2` is the side adjacent to `theta`, `x` is the hypotenuse and then the side opposite theta is `sqrt(4-x^2)`.

Then, `csctheta=x/sqrt(4-x^2)` and `cottheta=2/sqrt(4-x^2)`. So:

`I=-1/2lnabs(x/sqrt(4-x^2)+2/sqrt(4-x^2))=1/2lnabs(sqrt(4-x^2)/(x+2))`

Expanding further:

`I=1/4lnabs(4-x^2)-1/2lnabs(x+2)=1/4lnabs(2+x)+1/4lnabs(2-x)-1/2lnabs(x+2)`

`I=1/4lnabs(2-x)-1/4lnabs(x+2)`

We can combine these and switch the order of `2-x` to `x-2` since we're working with absolute values:

`I=1/4lnabs((x-2)/(x+2))+C`

This would be much quicker if done with partial fractions.

Answer 2

`int dx/(x^2-4) = 1/4 ln abs((x-2)/(x+2)) +C`

Explanation:

We can also show how to solve with partial fractions:

`int dx/(x^2-4) = int dx/((x-2)(x+2))`

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

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

`1= Ax+2A+Bx-2B`

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

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

`{(A = -B ),(4A =1):}`

`{(A=1/4),(B=-1/4):}`

`int dx/(x^2-4) = int (1/(4(x-2)) - 1/(4(x+2)))dx`

`int dx/(x^2-4) = 1/4 int dx/(x-2) - 1/4 int dx/(x+2)`

`int dx/(x^2-4) = 1/4 lnabs(x-2) - 1/4 ln abs(x+2) +C`

`int dx/(x^2-4) = 1/4 lnabs(x-2) - 1/4 ln abs(x+2) +C`

`int dx/(x^2-4) = 1/4 ln abs((x-2)/(x+2)) +C`