Partial Fractions for int [ (2x^2) / [(x - 3)(x + 2)^2] ] dx

[scrum]

I try to use cover up method and I've tried

my wrong answer is 18ln(x-3)+1/(x-2)+8ln(x-2)+C

 

[soroban]

Re: Partial Fractions

Hello,scrum!

It's hard to see your work from here,
. . but it looks like you played the \(\displaystyle J\heartsuit\) instead of the \(\displaystyle 7\spadesuit.\)

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

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

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

\(\displaystyle \text{Let }x = 3:\;\;18 \;=\;25A\quad\Rightarrow\quad\boxed{A \:=\:\frac{18}{25}}\)

\(\displaystyle \text{Let }x = -2:\;\;8 \:=\:-5C\quad\Rightarrow\quad\boxed{C \:=\:-\frac{8}{5}}\)

\(\displaystyle \text{Let }x = 0:\;\;0 \:=\:4A - 6B - 3C\)
. . \(\displaystyle 0 \:=\:4\left(\frac{18}{25}\right) - 6B - 3\left(-\frac{8}{5}\right)\quad\Rightarrow\quad\boxed{B \:=\:\frac{32}{25}}\)


\(\displaystyle \text{Therefore: }\;\frac{2x^2}{(x-3)(x+2)^2} \;=\;\frac{\frac{18}{25}}{x-3} + \frac{\frac{32}{25}}{x+2} - \frac{\frac{8}{5}}{(x+2)^2}\)


Try it again . . .

 

[Dr. Flim-Flam]

Scrum if you have a TI-89, to avoid unnecessary grunt work, just go to expand (F2-3) .

For example: expand(2x^2/((x-3)*(x+2)^2),x) = 32/[25(x+2)] -8/[5(x+2)^2] +18/[25(x-3)].

Now, taking the integral is a piece of cake. Saves a lot of time, for what it is worth.