what is the integral of (6x-2)/(x^2+1x-6)?

Answer 1

Partial fractions.

The denominator is equal to (x-2)(x+3).

So write:

(6x-2)/(x-2)(x+3) = a/(x-2) + b/(x+3) =[ a(x+3) + b(x-2)]/(x-2)(x+3)

So a(x+3) + b(x-2) = 6x - 2

So a+b = 6, and 3a - 2b = -2.

Solving for a and b gives a=2, b=4.

So (6x-2)/(x^2+x-6) = 2/(x-2) + 4/(x+3)

The integral of the right side is:

2 ln(x-2) + 4 ln(x+3) + C

Answer 2

first convert (6x-2)/(x^2+x-6) into a/(x+3) + b/(x-2)

(6x -2)/(x62 +x -6) = (a(x-2) + b(x+3))/((x+3)(x-2))
= (ax -2a +bx +3b)/((x+3)(x-2))
= (a+b)x + (-2a+3b)/((x+3)(x-2))

compare coeff. a+b = 6, and -2a+3b = -2
a = 4 b = 2

so the integral is 4ln(x-3) + 2ln(x-2)