How do you integrate #(7x-2)/((x-3)^2(x+1))# using partial fractions?

1 Answer
Nov 3, 2017

Write the partial fractions equation:

#(7x-2)/((x-3)^2(x+1)) = A/(x+1)+B/(x-3)+C/(x-3)^2#

Multiply both sides by #(x-3)^2(x+1)#:

#7x-2 = A(x-3)^2+B(x-3)(x+1)+C(x+1)#

Let #x = -1#

#7(-1)-2 = A(-1-3)^2+B(-1-3)(-1+1)+C(-1+1)#

#-9 = 16A#

#A = -9/16#

#7x-2 = -9/16(x-3)^2+B(x-3)(x+1)+C(x+1)#

Let #x = 3#

#7(3)-2 = -9/16(3-3)^2+B(3-3)(3+1)+C(3+1)#

#C = 19/4

#7x-2 = -9/16(x-3)^2+B(x-3)(x+1)+19/4(x+1)#

Let #x = 0#

#7(0)-2 = -9/16(0-3)^2+B(0-3)(0+1)+19/4(0+1)#

#-2 = -81/16-3B+19/4(0+1)#

#B = 9/16#

The partial fractions are:

#(7x-2)/((x-3)^2(x+1)) = -9/16 1/(x+1)+9/16 1/(x-3)+19/4 1/(x-3)^2#

Integrate:

#int (7x-2)/((x-3)^2(x+1)) dx = -9/16 int 1/(x+1) dx+9/16 int 1/(x-3) dx +19/4 int 1/(x-3)^2 dx#

The integrals are trivial:

#int (7x-2)/((x-3)^2(x+1)) dx = -9/16 lnt|x+1|+9/16 ln|x-3| -19/4 1/(x-3)+ C#