Evaluate int \ 1/(x^2-2x) \ dx ?
1 Answer
Mar 17, 2017
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|
