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}$???????