1. ## Partial fraction integration

$\displaystyle \int\frac{6x^{2}-12x+27}{(x-4)(x^{2}+9)} = \frac{A}{x-4} + \frac{B x + C}{x^{2}+9}$

Am I doing something wrong here?

First, I simplify this
$\displaystyle 3\int\frac{2x^{2}-4x+9}{(x-4)(x^{2}+9)}$

$\displaystyle \frac{A}{x-4}+\frac{BX+C}{x^2+9}$

is this setup correct?

$\displaystyle A(x^2+9)+(BX+C)(x-4)=2x^{2}-4x+9$

When I set x=4, I get an A value of 25/19

This does not seem correct. My prof said we would ONLY get whole numbers for the example problems given.

What should I do differently?

Thank you!

2. $\displaystyle f(4) : A(4^2+9) = 32-16+9 \implies 25A = 25 \implies A=1$

3. Originally Posted by Vamz
$\displaystyle \int\frac{6x^{2}-12x+27}{(x-4)(x^{2}+9)} = \frac{A}{x-4} + \frac{B x + C}{x^{2}+9}$

Am I doing something wrong here?

First, I simplify this
$\displaystyle 3\int\frac{2x^{2}-4x+9}{(x-4)(x^{2}+9)}$

$\displaystyle \frac{A}{x-4}+\frac{BX+C}{x^2+9}$

is this setup correct?

$\displaystyle A(x^2+9)+(BX+C)(x-4)=2x^{2}-4x+9$

When I set x=4, I get an A value of 25/19

This does not seem correct. My prof said we would ONLY get whole numbers for the example problems given.

What should I do differently?

Thank you!
Since you didn't show your work I can only guess that you have made an arithmetic error
$\displaystyle \displaystyle x=4 \implies A(4^2+9)+(Bx+c)(4-4)=2(4)^2-4(4)+9 \iff 25A=25 \implies A=1$