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

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Answer 1 · 2016-06-29T19:48:10

$= ln \ (x-1)/(x+2) + C$

$int dx qquad 3/(x^2+x-2)$

well $x^2+x-2 = (x+2)(x-1)$

so we can say that

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

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

$\implies A(x-1) + B(x+2) = 3$

let x = 1 so B*3 = 3, so B = 1

let x = -2 so A*(-3) = 3 so A = -1

so the integral becomes

$int dx qquad 1/(x-1) - 1/(x+2)$

$= ln(x-1) - ln (x+2) + C$

$= ln \ (x-1)/(x+2) + C$

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