Greetings everyone. I'm trying to do the following proof my own way but it doesn't look like my final answer is matching the solution's. Is there something wrong?

The solution shows the result by picking specific numbers of $\displaystyle a, b, c $ and doing some tricky algebra, right? But I want to avoid the tricky algebra

so I'm trying to use long division to "pull out" or derive the integration constant.

Thank you everyone.

$\displaystyle \text{Given that P(}x)=Q(x)R'(x)-Q'(x)R(x),\text{ write down an expression for } \int \frac{P}{Q^2} dx. $

$\displaystyle \text{By choosing } R(x)=a+bx+cx^2, \text{show}$

$\displaystyle \int \frac{5x^2 - 4x - 3}{(1 + 2x + 3x^2)^2} dx = \frac{\frac{2+c}{3}+ \frac{2c-5}{3}x+cx^2}{1+2x+3x^2}.$

$\displaystyle \text{Since the choice of }R(x)\text{ is not unique, by comparing the two functions }R(x) $

$\displaystyle \text{corresponding to two different values of }a,\text{ explain how the different choices are related}\text{.} $

Solution:

My work(My latex is below but I didn't get it right)

$\displaystyle 3{{x}^{2}}+2x+1\overset{\frac{c}{3}}{\overline{\le ft){\begin{align}$$\displaystyle

& \text{ }c{{x}^{2}}+\frac{2c-5}{3}x+\frac{2+c}{3} \\

& \underline{-\left( c{{x}^{2}}\text{ }+\text{ }\frac{2c}{3}x+\text{ }\frac{c}{3} \right)} \\

& \text{ }\frac{-5}{3}x\text{ + }\frac{2}{3} \\

\end{align}}\right.}}\text{ so }\frac{R(x)}{Q(x)}=\frac{c}{3}+\frac{1}{3}\left( \frac{-5x+2}{1+2x+3{{x}^{2}}} \right) \\

& \text{But as you can see above, solution has }\frac{R(x)}{Q(x)}=\text{constant + }\frac{-3x-2{{x}^{2}}}{1+2x+3{{x}^{2}}} \\

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