Question

Evaluate `int \ 1/(x^2-2x) \ dx`?

Answer 1

`int \ 1/(x^2-2x) \ dx = 1/2 ln |(A(x-2))/x|`

Explanation:

We want to find:

`int \ 1/(x^2-2x) \ dx`

If we examine the integrand we can decompose into partial fractions, as follows:

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

And so:

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

Put:

`x = 0=> 1 = -2A => A = -1/2`
`x = 2=> 1 = 2B \ \ \ \ \=> B = 1/2`

Therefore:

`int \ 1/(x^2-2x) \ dx = int \ (1/2)/(x-2) - (1/2)/x \ dx`
`" " = 1/2 ln |x-2| - 1/2ln|x| + C`
`" " = 1/2 ln |(x-2)/x| + C`
`" " = 1/2 ln |(A(x-2))/x|`