# Thread: Integration of x/x-6 dx , by partial fractions

1. ## Integration of x/x-6 dx , by partial fractions

The question is to evaluate the integral x/x-6 dx by partial fractions

This is how I've started

$
\begin{array}{l}
\int {\frac{x}{{x - 6}}dx} \\
= \int {\frac{{x - 1 + 1}}{{x - 6}}dx = \int {\frac{{x - 1}}{{x - 6}} + \frac{1}{{x - 6}}} } \\
\end{array}
$

so giving

$
\begin{array}{l}
= \int {\frac{{x - 1 + 1}}{{x - 6}}dx = \int {\frac{{x - 1}}{{x - 6}} + \frac{1}{{x - 6}}} } \\
\frac{A}{{x - 6}} + \frac{B}{{x - 6}} = \frac{{x - 1 + 1}}{{x - 6}} \\
A(x - 6)(x - 6) + B(x - 6)(x - 6) = x(x - 6) \\
A(x^2 - 12x + 36) + B(x^2 - 12 + 36) = x^2 - 6x \\
\end{array}
$

rearranging this to

$
(A + B)x^2 + ( - 12A - 12B)x + (A + B)36 = x^2 - 6x
$

and then I have

A+B = 1 due to the co-efficient of x^2 term of polynomial both sides
-12A-12B= -6 due to the co-efficient of x
and A+B = 0 due to the constant

I'm really having trouble on how to solve for these .

2. Originally Posted by Craka
The question is to evaluate the integral x/x-6 dx by partial fractions

This is how I've started

$
\begin{array}{l}
\int {\frac{x}{{x - 6}}dx} \\
= \int {\frac{{x - 1 + 1}}{{x - 6}}dx = \int {\frac{{x - 1}}{{x - 6}} + \frac{1}{{x - 6}}} } \\
\end{array}
$

so giving

$
\begin{array}{l}
= \int {\frac{{x - 1 + 1}}{{x - 6}}dx = \int {\frac{{x - 1}}{{x - 6}} + \frac{1}{{x - 6}}} } \\
\frac{A}{{x - 6}} + \frac{B}{{x - 6}} = \frac{{x - 1 + 1}}{{x - 6}} \\
A(x - 6)(x - 6) + B(x - 6)(x - 6) = x(x - 6) \\
A(x^2 - 12x + 36) + B(x^2 - 12 + 36) = x^2 - 6x \\
\end{array}
$

rearranging this to

$
(A + B)x^2 + ( - 12A - 12B)x + (A + B)36 = x^2 - 6x
$

and then I have

A+B = 1 due to the co-efficient of x^2 term of polynomial both sides
-12A-12B= -6 due to the co-efficient of x
and A+B = 0 due to the constant

I'm really having trouble on how to solve for these .
Maybe you have miswritten the question but

$\frac{x}{x-6}=\frac{x-6+6}{x-6}=1+\frac{6}{x-6}$ You dont perform P-fracs on a rational function where the numerator and denominator are of equal degree.

3. I've checked that is the question.
I know normally you use partial fractions when the degree of the numerator is less than that of the denominator, but this is one of the questions in the text to be done by partial fractions.

4. Originally Posted by Craka
I've checked that is the question.
I know normally you use partial fractions when the degree of the numerator is less than that of the denominator, but this is one of the questions in the text to be done by partial fractions.
P-fracs doesnt work when the numerator and denominator are of equal degree though...

5. Ok, rotten text.

Guess will have to work out by substitution than.

Thanks