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

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Answer 1 · 2016-09-14T18:20:53

#ln|x-1|+2ln|x+4|+C#,

or,

#ln|(x-1)(x+4)^2|+C#.

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

To decompose the integrand using Method of Partial Fraction, let,

#(3x+2)/((x-1)(x+4))=A/(x-1)+B/(x+4)", where, "A,B in RR#.

We use Heavyside's Cover-up Method to determine #A, and, B# :

#A=[(3x+2)/(x+4)]_(x=1) =(3+2)/(1+4)=1#

#B=[(3x+2)/(x-1)]_(x=-4) =(-12+2)/(-4-1)=2#. Hence,

#I=int[1/(x-1)+2/(x+4)]dx#

#=int1/(x-1)dx+2int1/(x+4)dx#

#=ln|x-1|+2ln|x+4|#

Therefore,

#I=ln|x-1|+2ln|x+4+C|#, or,

#I=ln|(x-1)(x+4)^2|+C#.

Enjoy Maths.!