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!

 A: Recall that
\begin{align}
\frac1{a-b z} &=\frac1a\frac1{1-\left(\frac b a z\right)}\\
&=\frac1a\sum_{n=0}^{\infty}\left(\frac b a z\right)^n
\end{align}

We have $$\frac1{1+2x}=\frac1{1-(-2x)}=1-2x+4x^2-8x^2\pm \ldots$$ and $$\frac1{2-3x}=\frac12\frac1{1-\frac32x}=\frac12\times\left(1+\frac32x+\frac94x^2\pm\ldots\right)$$
Now you have $1+6\times\frac1{1+2x} + 5 \times \frac1{2-3x}$, $\color{red}{\mbox{and you can substitute the latter expansions and obtain the result you are after.}}$
A: 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 $\frac{237}{8}$.
