Here is what I've done so far...

$\displaystyle (x^3-4x^2-16)/((x^2)(x-4))=A/x+B/x^2+C/(x-4)$

so

$\displaystyle x^3-4x^2-16=A(x^2)(x-4)+B(x)(x-4)+C(x)(x^2)$

=$\displaystyle A(x^3-4x^2)+B(x^2-4x)+C(x^3)$

=$\displaystyle (A+C)(x^3)+(-4A+B)(x^2)+(-4B)(x)$

Now we have to find what the coefficients are equal to, so

A+C=1

-4A+B=-4

B=0

A=1

B=0

C=0

So now I should be able to take the integral of $\displaystyle 1/x+0/x^2+0/(x-4)$

This equals $\displaystyle ln(x)$ on the interval [7,8], so my answer should be $\displaystyle ln(8)-ln(7)$.

This is apparently the wrong answer, so I must have made a mistake somewhere, but I'm not seeing it.