# Evaluate the integral of (x - 9)/(x^2 - 3x - 18) dx, using the method of partial fraction...

## Question:

Evaluate the integral {eq}\int \frac{x - 9}{x^2 - 3x - 18} \, \mathrm{d}x {/eq}, using the method of partial fraction decomposition.

## Integration by Partial Fraction:

{eq}\\ {/eq}

We know that a rational function is the function of the form {eq}\dfrac{p(x)}{q(x)} {/eq}, where {eq}q(x)\neq 0 {/eq}. If the degree of numerator is less than the degree of denominator, then the fraction is said to be proper, otherwise it is an improper fraction.

The partial fraction conversion used in this question is stated below :-

{eq}\dfrac{px+q}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b} \ , \ a\neq b {/eq}

{eq}\\ {/eq}

Using partial decomposition, the integral {eq}\displaystyle\int \frac{x - 9}{x^2 - 3x - 18} \, \mathrm{d}x {/eq} can also be written as :-

{eq}\displaystyle\int \frac{x - 9}{x^2 - 6x+3x - 18} \, \mathrm{d}x\\ \Rightarrow \displaystyle\int \frac{x - 9}{(x-3)(x - 6)} \, \mathrm{d}x \\\Rightarrow \displaystyle\int \frac{x - 3-6}{(x-3)(x - 6)} \, \mathrm{d}x\\\Rightarrow \displaystyle\int \frac{1}{x - 6} \, \mathrm{d}x-\displaystyle\int \frac{6}{(x-3)(x - 6)} \, \mathrm{d}x\\\Rightarrow \displaystyle\int \frac{1}{x - 6} \, \mathrm{d}x+2\displaystyle\int \frac{1}{x-3} \, \mathrm{d}x-2\displaystyle\int\frac{1}{x-6} \, \mathrm{d}x\\\Rightarrow \log|x-6|+2\log|x-3|-2\log|x-6|+C {/eq}