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.!

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