# Partial Fractions

• Mar 11th 2009, 10:10 PM
hlpplz
Partial Fractions
∫(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?
• Mar 11th 2009, 10:27 PM
Jhevon
Quote:

Originally Posted by hlpplz
∫(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?

$\frac {4x^2 - 11x + 4}{x^3 - 3x^2 + 2x} = \frac {4x^2 - 11x + 4}{x(x - 2)(x - 1)} = \frac Ax + \frac B{x - 2} + \frac C{x - 1}$

multiplying both sides of the equation by the LCD, we have:

$4x^2 - 11x + 4 = A(x - 2)(x - 1) + Bx(x - 1) + Cx(x - 2)$

now can you finish?
• Mar 11th 2009, 10:27 PM
Quote:

Originally Posted by hlpplz
∫(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?

∫(4x²-11x+4)/(x³-3x²+2x)

$= \int{\frac{A}{x} +\frac{B}{x-2} +\frac{C}{x-1}}$

Find A B & C And then integrate

its too late (Giggle)
• Mar 11th 2009, 10:43 PM
hlpplz
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?
• Mar 11th 2009, 10:46 PM
Quote:

Originally Posted by hlpplz
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⎮ +c

Good?

Yes(Clapping)
• Mar 11th 2009, 10:49 PM
hlpplz
Thank you very much (Rock)