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| #
