How do you integrate #int (4x - 16) / (x^2 - 2x - 3)# using partial fractions?

How do you integrate #int (4x - 16) / (x^2 - 2x - 3)dx# using partial fractions?

1 Answer
Sep 4, 2016

#-ln|x-3|+5ln|x+1|+C#,

OR,

#ln|(x+1)^5/(x-3)|+C#.

Explanation:

Let #I=int(4x-16)/(x^2-2x-3)dx=4int(x-4)/((x-3)(x+1))dx#

We decompose the Integrand

#(x-4)/((x-3)(x+1)) = A/(x-3)+B/(x+1)," where, "A,B in RR#.

Using Heavyside's Cover-up Method to find consts. #A,B in RR#,

#A=[(x-4)/(x+1)]_(x=3) =(3-4)/(3+1)=-1/4#,

#B=[(x-4)/(x-3)]_(x=-1)=(-1-4)/(-1-3)=5/4#.

Hence, #I=4int[(-1/4)/(x-3)+(5/4)/(x+1)]dx#

#=-int1/(x-3)dx+5int1/(x+1)dx#

#:. I=-ln|x-3|+5ln|x+1|+C#, or,

#I=ln|(x+1)^5/(x-3)|+C#.

Enjoy Maths.!