# How do you find int (x^2 - 5x) / ((x-1)(x+1)(x-2)) dx using partial fractions?

Apr 18, 2016

$\setminus \int \setminus \frac{{x}^{2} - 5 x}{\left(x - 1\right) \left(x + 1\right) \left(x - 2\right)} = 2 \ln | x - 1 | + \ln | x + 1 | - 2 \ln | x - 2 | + C$

#### Explanation:

$\setminus \frac{{x}^{2} - 5 x}{\left(x - 1\right) \left(x + 1\right) \left(x - 2\right)} = \setminus \frac{A}{x - 1} + \setminus \frac{B}{x + 1} + \setminus \frac{C}{x - 2}$

Multiplying both sides by the common denominator:

${x}^{2} - 5 x = A \left(x + 1\right) \left(x - 2\right) + B \left(x - 1\right) \left(x - 2\right) + C \left(x - 1\right) \left(x + 1\right)$

${x}^{2} - 5 x = A \left({x}^{2} - x - 2\right) + B \left({x}^{2} - 3 x + 2\right) + C \left({x}^{2} - 1\right)$

${x}^{2} - 5 x = A {x}^{2} - A x - 2 A + B {x}^{2} - 3 B x + 2 B + C {x}^{2} - C$

$\left(A {x}^{2} + B {x}^{2} + C {x}^{2} - {x}^{2}\right) + \left(- A x - 3 B x + 5 x\right) + \left(- 2 A + 2 B - C\right) = 0$

$\left(A + B + C - 1\right) {x}^{2} + \left(- A - 3 B + 5\right) x + \left(- 2 A + 2 B - C\right) = 0$

So now, you have 3 system of equations:

$A + B + C = 1$
$- A - 3 B = - 5$
$- 2 A + 2 B - C = 0$

Solving for A, B and C, you'll get:
$A = 2 , B = 1 , C = - 2$

Substitute these back to the equation:
$\setminus \frac{{x}^{2} - 5 x}{\left(x - 1\right) \left(x + 1\right) \left(x - 2\right)} = \setminus \frac{2}{x - 1} + \setminus \frac{1}{x + 1} + \setminus \frac{- 2}{x - 2}$

Now you can easily get the Integral of each fraction:
$\setminus \int \setminus \frac{2}{x - 1} + \setminus \int \setminus \frac{1}{x + 1} + \setminus \int \setminus \frac{- 2}{x - 2}$

$= 2 \ln | x - 1 | + \ln | x + 1 | - 2 \ln | x - 2 | + C$