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#