# partial fractions

• Mar 26th 2009, 05:19 PM
qzno
partial fractions
$\int \frac{3x^5 - 7x^4 - 27x^3 + 28x^2 + 16}{x^3 - 4x^2} dx$

where the degree of the numerator is higher than the denominator, i thought i had to do polynomial long division so i got:

$
\int 3x^2 + 5x - 7 + \frac{16}{x^3-4x^2} dx$

and im now at this point and am stuck

$
x^3 + \frac{5x^2}{2} - 7x + 16\int \frac{1}{x^2(x-4)} dx$
• Mar 26th 2009, 06:11 PM
skeeter
put the 16 back.

repeated linear factor ...

$\frac{16}{x^2(x-4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-4}$

$A = -1$ , $B = -4$ , $C = 1$