# Thread: partial fractions

1. ## partial fractions

Hia, I tried using partial fractions to integrate this expression but the integrator function on wolfram mathematics gave me a different answer.

$\displaystyle \int \frac{x^3}{(x^2-9)(x+1)}dx$ $\displaystyle =\int \frac{x^3}{(x+3)(x-3)(x+1)}dx$

integrals.wolfram.com gave me $\displaystyle \frac{1}{8}(8x+9ln(x-3)+ln(x+1)-18ln(x+3))$ where I presume the x is another constant

whilst I get $\displaystyle \frac{7}{4}ln(x+3)+\frac{9}{8}ln(x-3)+\frac{1}{8}ln(x+1)$ which is almost there,,,but if anyone could help me that would be great! thankyou
For my partial fractions I substituted x=-1,x=3 then when I did x=1 to calculate the coefficient above the term x+3 I kept getting 7/4. It must be a minor mistake which I can't see anywhere!!!!!
$\displaystyle \frac{x^3}{(x+1)(x-3)(x+3)}=\frac{A}{x+3}+\frac{B}{x-3}+\frac{C}{x+1}$
$\displaystyle x^3=A(X-3)(X+1)+B(X+3)(X+1)+C(x+3)(x-3)$
x=1$\displaystyle \Longrightarrow$$\displaystyle C=\frac{1}{8}X=3\displaystyle \Longrightarrow$$\displaystyle B=\frac{27}{24}$X=1$\displaystyle \Longrightarrow$$\displaystyle A=\frac{7}{4}$???????

2. Hello, beaux!

You overlooked something quite important. . .

$\displaystyle \int \frac{x^3}{(x+3)(x-3)(x+1)}dx$

The degree of of the numerator is not less than the degree of the denominator.
. . The fraction is "improper" (top-heavy).

We must use Long Division . . .

$\displaystyle x^3 \div(x^3 + x^2-9x-9) \;=\;1 - \underbrace{\frac{x^2-9x-9}{(x-3)(x+3)(x+1)}}$

. . . . . .$\displaystyle \text{and apply Partial Fractions to }\overbrace{\text{this}}^{\uparrow}\text{ fraction.}$

3. hia again,i dont understand?! Both the numerator and denominator are cubic but the denominator also has a linear expression with x also. Does this not make the denominator's degree bigger than the numerators?ie proper?