I got an answer but I'm not sure if i did the problem correctly.
Evaluate the following integral using partial fractions:
The integral of (x+1)/(x^2 - 5x + 6)
My answer was -3ln(x-2)+(4/3)ln(x-3) + C
Is this correct???
Almost! the coefficient of ln(x - 3) is just 4, not 4/3
(x+1)/(x^2 - 5x + 6) = (x+1)/(x - 3)(x - 2)
so we have (x+1)/(x - 3)(x - 2) = A/(x - 3) + B/(x - 2)
multiply through by (x - 3)(x - 2) we obtain
x + 1 = A(x - 2) + B(x - 3)
=> x + 1 = Ax - 2A + Bx - 3B
=> x + 1 = (A + B)x + (-2A - 3B)
=> A + B = 1..........................(1)
and -2A - 3B = 1 ....................(2)
2A + 2B = 2 ................(3) = (1)*2
-2A - 3B = 1 ...............(2)
=> -B = 3
=> B = -3
=> A - 3 = 1
=> A = 4
so (x+1)/(x^2 - 5x + 6) = 4/(x - 3) - 3/(x -2)
so int{(x+1)/(x^2 - 5x + 6)}dx = int{4/(x - 3) - 3/(x -2)}dx
= 4ln(x - 3) - 3ln(x - 2) + C ...........so you were almost correct.
Remember, whenever you integrate and get an answer, you can check if you're right by differentiating the answer you got.
d/dx(4ln(x - 3) - 3ln(x - 2) + C) = 4*(1/x - 3) - 3*(1/(x - 2))
= 4/(x - 3) - 3/(x - 2) + 0
= [4(x - 2) - 3(x - 3)]/[(x - 3)(x - 2)]
= (4x - 8 - 3x + 9)/(x - 3)(x - 2)
= (x + 1)/(x - 3)(x - 2) ................as desired