# How do you integrate (x-5)/(x-2)^2 using partial fractions?

Jul 29, 2016

$\int \frac{x - 5}{x - 2} ^ 2 . d x = l n \left(| x - 2 |\right) - \frac{3}{x - 2} + C$

#### Explanation:

int (x-5)/(x-2)^2 .d x=?

$\text{let us write (x-5) as (x-2-3)}$

$\int \frac{x - 2 - 3}{x - 2} ^ 2 \cdot d x$

"split (x-2-3 )

$\int \frac{\cancel{\left(x - 2\right)}}{\cancel{{\left(x - 2\right)}^{2}}} \cdot d x - \int \frac{3}{x - 2} ^ 2 \cdot d x$

$\int \frac{d x}{x - 2} - 3 \int \frac{d x}{x - 2} ^ 2$

$\text{1- solve } \int \frac{d x}{x - 2}$

$\text{substitute u=x-2" ; } d u = d x$

$\int \frac{d x}{x - 2} = \frac{d u}{u} = l n u$

$\text{undo substitution }$

$\int \frac{d x}{x - 2} = l n \left(x - 2\right)$

$\text{2- now solve } 3 \int \frac{d x}{x - 2} ^ 2$

$u = x - 2 \text{ ; " d x=d u" ; } 3 \int \frac{d u}{u} ^ 2 = 3 \int {u}^{-} 2 d u$

$3 \int \frac{d x}{x - 2} = \frac{1}{- 2 + 1} {u}^{- 2 + 1} = - 3 \cdot {u}^{- 1} = - \frac{3}{u}$

$\text{undo substitution}$

$3 \int \frac{d x}{x - 2} ^ 2 = - \frac{3}{x - 2}$

$\text{The problem is solved:}$

$\int \frac{x - 5}{x - 2} ^ 2 . d x = l n \left(| x - 2 |\right) - \frac{3}{x - 2} + C$