∫(4x²-11x+4)/(x³-3x²+2x) dx
i know the bottom part breaks up into
x(x²-3x+2) = x(x-2)(x-1)
where do i go form there?
$\displaystyle \frac {4x^2 - 11x + 4}{x^3 - 3x^2 + 2x} = \frac {4x^2 - 11x + 4}{x(x - 2)(x - 1)} = \frac Ax + \frac B{x - 2} + \frac C{x - 1}$
multiplying both sides of the equation by the LCD, we have:
$\displaystyle 4x^2 - 11x + 4 = A(x - 2)(x - 1) + Bx(x - 1) + Cx(x - 2)$
now can you finish?
ok so i multiplied by the LCD and got
4x²-11x+4 = A(x-2)(x-1) + B(x)(x-1) + C(x)(x-2)
4x²-11x+4 = A(x²-3x+2) + B(x²-x) + C(x²-2x)
isolating x² = A + B + C = 4
isolating x = -3A - B - 2C = -11
isolating # = 2A = 4
solving for A,B,C i got
A= 2, B=-7, C=9
∫ A/x + B/(x-2) + C/(x-1) dx
∫ 2/x - 7/(x-2) + 9/(x-1) dx
2∫ 1/x - 7∫ 1/(x-2) + 9∫ 1/(x-1) dx
2ln⎮x⎮ - 7ln⎮x-2⎮ + 9ln⎮x-1⎮
Good?