How do you integrate $int (3x^2 + 10x -5)/ ( (x+1)^2(x-2) )$ using partial fractions?

Question

1951 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-19T10:56:05

$int(3x^2+10x-5)/((x+1)^2(x-2))dx=-4/(x+1)+3ln|x-2| +C$

Let's start with :

$(3x^2+10x-5)/((x+1)^2(x-2))=A/(x+1)+B/(x+1)^2+C/(x-2)=$

$(A(x+1)(x-2)+B(x-2)+C(x+1)^2)/((x+1)^2(x-2))=$

$(A(x^2-x-2)+Bx-2B+C(x^2+2x+1))/((x+1)^2(x-2))=$

$(Ax^2-Ax-2A+Bx-2B+Cx^2+2Cx+C)/((x+1)^2(x-2))=$

$((A+C)x^2+(-A+B+2C)x+(-2A-2B+C))/((x+1)^2(x-2))$

So we know now :

$A+C=3$
$-A+B+2C=10$
$-2A-2B+C=-5$

From this system we get :

$A=0, B=4, C=3$

Let's get to the integral now:

$int(3x^2+10x-5)/((x+1)^2(x-2))dx=int(4/(x+1)^2+3/(x-2))dx=$

$int4/(x+1)^2dx+int3/(x-2)dx=$

$int4(x+1)^(-2)dx+3ln|x-2|=$

$-4(x+1)^(-1)+3ln|x-2|=$

$-4/(x+1)+3ln|x-2| +C$

How do you integrate int (3x^2 + 10x -5)/ ( (x+1)^2(x-2) )int (3x^2 + 10x -5)/ ( (x+1)^2(x-2) ) using partial fractions?