<s>[INST] Obtain the coefficient of $x^2$ in the expansion of $1+\frac{6}{2x+1}+\frac{5}{2-3x}$ Hello so this is a 2 part question and I managed to express that praction as a partial fraction which was equaled to $$1+\frac{6}{2x+1}+\frac{5}{2-3x}$$ I will add my work below I tried lot to Obtain the coefficient of $x^2$ but cant get to the answer of 237/8 and help is much appreciated. Thank you!

 [/INST] This is how to do it the hard way.
Let's suppose that, for $x$ close enough to $0$,
$$1 + \frac{6}{2x+1} + \frac{5}{2-3x} = a_0 + a_1x + a_2x^2 + \dots$$
$$(2x+1)(2-3x)\left(1 + \frac{6}{2x+1} + \frac{5}{2-3x}\right) = 19 - 7x - 6x^2 + \dots$$
$$(2x+1)(2-3x)(a_0 + a_1x + a_2x^2) 
   = 2a_0 + (a_0+2a_1)x + (-6 a_0 + a_1 + 2 a_2)x^2 + \dots$$
Comparing $19 - 7x - 6x^2$  to $2a_0 + (a_0+2a_1)x + (-6 a_0 + a_1 + 2 a_2)x^2$
\begin{align}
   2a_0 &= 19 \\
   a_0 &= \frac{19}{2} \\
\hline
   a_0 + 2a_1 &= -7 \\
   \frac{19}{2} + 2a_1 &= -7 \\
   a_1 &= -\frac{33}{4} \\
\hline
   -6 a_0 + a_1 + 2 a_2 &= -6 \\
   -57 - \frac{33}{4} + 2a_2 &= -6 \\
   a_2 &= \frac{237}{8}
\end{align}
So the coefficient of x^2 is $\frac{237}{8}$.
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