# How do you integrate int (2x - 1) / [(x - 1)^3 (x - 2)] using partial fractions?

Jul 9, 2017

I decomposed integrand into basic fractions,

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

$\left(2 x - 1\right) = A \cdot \left({x}^{3} - 4 {x}^{2} + 5 x - 2\right) + B \cdot \left({x}^{2} - 3 x + 2\right) + C \cdot \left(x - 2\right) + D \cdot \left({x}^{3} - 3 {x}^{2} + 3 x - 1\right)$

Let $x = 1$, $- C = 1$ or $C = - 1$.

Let $x = 2$, $D = 3$.

Differentiate both sides,

$2 = A \cdot \left(3 {x}^{2} - 8 x + 5\right) + B \cdot \left(2 x - 3\right) + C + D \cdot \left(3 {x}^{2} - 6 x + 3\right)$

Let $x = 1$, $- B + C = 2$ or $- B - 1 = 2$. Hence $B = - 3$

Differentiate both sides,

$0 , = A \cdot \left(6 x - 8\right) + 2 B + D \cdot \left(6 x - 6\right)$

Let $x = 1$, $- 2 A + 2 B = 0$ or $A = B = - 3$.

Thus,

$\int \frac{2 x - 1}{{\left(x - 1\right)}^{3} \cdot \left(x - 2\right)} \mathrm{dx}$

$= \int - \frac{3}{x - 1} \mathrm{dx} + \int - \frac{3}{x - 1} ^ 2 \mathrm{dx} + + \int - \frac{1}{x - 1} ^ 3 \mathrm{dx} + \int \frac{3}{x - 2} \mathrm{dx}$

$= - 3 L n \left(x - 1\right) + \frac{3}{x - 1} + \frac{1}{2} \cdot {\left(x - 1\right)}^{- 2} + 3 L n \left(x - 2\right) + C$

$= \frac{3}{x - 1} + \frac{1}{2} \cdot {\left(x - 1\right)}^{- 2} + 3 L n \left(\frac{x - 2}{x - 1}\right) + C$

#### Explanation:

I decomposed integrand into basic fractions.