How do you integrate #int (4x+3) / (x - 1)^2dx# using partial fractions?

1 Answer
May 22, 2017

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Explanation:

#int (4(x-1)+4+3)/(x-1)^2 dx#

By substituting #x# in #4x# with the denominator, the fraction is split into partial fractions. Include a #+4# so the original fraction isn't changed.

#int (4cancel(x-1))/(x-1)^cancel2 +7/(x-1)^2dx #

Split the integrals to make it easier.

#=int 4/(x-1) dx + int 7/(x-1)^2 dx#

#=4 int 1/(x-1)dx + 7int(x-1)^-2dx#

  1. #4 int 1/(x-1)dx = 4ln|x-1| + c_1# Integration of inverse functions.

  2. #7 int(x-1)^-2dx = 7*-1*(x-1)^-1 + c_2# Using reverse chain rule.

Therefore,

#=4 int 1/(x-1)dx + 7int(x-1)^-2dx#

#=4ln|x-1| - 7(x-1)^-1 +c_3#