Question

How do you evaluate the integral `int 1/(x^2-2x-3)dx` from 0 to 4?

Answer 1

We seek:

`int_0^4 \ 1/(x^2-2x-3) \ dx = -1/4 \ (ln5+ln3)`

Explanation:

We seek:

`I = int_0^4 \ 1/(x^2-2x-3) \ dx`

We can factorise the integrand and decompose into partial fraction as follows:

`1/(x^2-2x-3) -= 1/((x-3)(x+1))`
`" " = A/(x-3) + B/(x+1)`
`" " = (A(x+1) + B(x-3))/((x-3)(x+1))`

Leading to the identity:

`1 = A(x+1) + B(x-3)`

Where `A,B` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x = +3 => 1 = 4A => A = 1/4`
Put `x = -1 => 1=-4B => B = -1/4`

Thus:

`I = int_0^4 \ (1/4)/(x-3) + (-1/4)/(x+1) \ dx`
`\ \ = 1/4 \ int_0^4 \ 1/(x-3) - 1/(x+1) \ dx`
`\ \ = 1/4 \ [ ln|x-3| - ln |x+1|]_0^4`
`\ \ = 1/4 \ {(ln1-ln5)-(ln3-ln1)}`
`\ \ = 1/4 \ {0-ln5-ln3-0}`
`\ \ = -1/4 \ (ln5+ln3)`