partial fractions--am i doing this right?

[missyc8]

integral(2x^2-9x-9/ x^3-9x)

so we use partial fractions because the bottom is bigger
bottom = x(x-3)(x+3)

A/ x + B/ x-3 + C/x+3 = A(x-3) + Bx(x+3) + Cx(x-3)

is this right? i get stuck after this...

 

[Deleted member 4993]

integral(2x^2-9x-9/ x^3-9x)

so we use partial fractions because the bottom is bigger
bottom = x(x-3)(x+3)

A/ x + B/ x-3 + C/x+3 = A(x-3) + Bx(x+3) + Cx(x-3)

is this right? No

i get stuck after this...

A/ x + B/ x-3 + C/x+3 = [A(x+3)(x-3) + Bx(x+3) + Cx(x-3)]/{x(x+3)(x-3)}

 

[soroban]

Hello, missyc8!

\(\displaystyle \int \frac{2x^2-9x-9}{x^3-9x}\,dx\)

\(\displaystyle \text{We have: }\;\frac{2x^2-9x-9}{x(x-3)(x+3)} \;\;=\;\;\frac{A}{x} + \frac{B}{x-3} + \frac{C}{x+3}\)

\(\displaystyle \text{Then: }\;2x^2 - 9x - 9 \;\;=\;\;(x-3)(x+3)A + x(x+3)B + x(x-3)C\)

\(\displaystyle \begin{array}{cccccc}\text{Let }x = 0\!: & 0 - 0 - 9 \;=\;(\text{-}3)(3)A + 0\cdot B + 0\cdot C & \Rightarrow & \text{-}9 \:=\:\text{-}9A & \Rightarrow& A \:=\:1 \\ \\[-4mm] \text{Let }x = 3\!: & 18 - 27 - 9 \;=\;0\cdot A + (3)(6)B + 0\cdot C & \Rightarrow & \text{-}18 \:=\:18B & \Rightarrow & B \:=\:\text{-}1 \\ \\[-4mm] \text{Let }x = \text{-}3\!: & 18 + 27 - 9 \;=\;0\cdot A + 0\cdot B + (\text{-}3)(\text{-}6)C & \Rightarrow & 36 \:=\:18C & \Rightarrow& C \:=\:2 \end{array}\)

\(\displaystyle \text{Therefore: }\;\frac{2x^2-9x-9}{x(x-3)(x+3)} \;=\;\frac{1}{x} - \frac{1}{x-3} + \frac{2}{x+3}\)

I assume you can finish it now . . .