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

Question

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

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Answer 1 · 2016-09-04T17:41:34

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

OR,

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

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.!