1. ## Partial Fractions

I cant solve this problem:

INTEGRAL OF: (x^3-4x^2-16) / (x^3-4x^2)

I change the denominator into (x^2)(x-4) and setup the following partial fraction:

A/x + (Bx+C)/x^2 + D/x-4

When you multiply the numerators by the other two denominators though you don't get a term without an x, thus your left with nothing to set equal to 16.

2. For partial fractions to work the degree of the numerator must be less than the degree of the denominator. Since both the numerator and denominator have a degree of three, you must do polynomial long division.

$\displaystyle \frac{{x^3 - 4x^2 - 16}} {{x^3 - 4x^2 }} = x^3 - 4x^2 \left){\vphantom{1{x^3 - 4x^2 - 16}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{x^3 - 4x^2 - 16}}}$

After this, take the answer from the long division plus the remainder and then perform the partial fraction decomposition.

3. $\displaystyle \frac{x^{3}-4x^{2}-16}{x^{3}-4x^{2}}=1-\frac{16}{x^{3}-4x^{2}}.$

The remaining challenge is to descompose $\displaystyle \frac{16}{x^{3}-4x^{2}},$ and we can tackle it like this:

$\displaystyle \frac{16}{x^{2}(x-4)}=\frac{x^{2}-(x-4)(x+4)}{x^{2}(x-4)}=\frac{1}{x-4}-\frac{x+4}{x^{2}}=\frac{1}{x-4}-\frac{1}{x}-\frac{4}{x^{2}}.$