How do you use partial fractions to find the integral #int (2x-3)/(x-1)^2dx#?

Question

2779 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-09T09:58:34

#int frac (2x-3) ((x-1)^2) dx = 2 ln|x-1| +1/(x-1)#

As the denominator of the integrand is already factorized we can quickly decompose it using partial fractions:

#frac (2x-3) ((x-1)^2) = A/(x-1) + B/((x-1)^2)#

#frac (2x-3) ((x-1)^2) = frac (A(x-1) + B) ((x-1)^2) = frac (Ax-A + B) ((x-1)^2)#

Hence:

#A=2#
#B=-1#

and:

#int frac (2x-3) ((x-1)^2) dx = 2 int dx/(x-1) - int dx/((x-1)^2)#

#int frac (2x-3) ((x-1)^2) dx = 2 ln|x-1| +1/(x-1)#