Answer

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**Hint:**To resolve the given term into partial fractions, we will be using a method of splitting the term into partial fractions with respect to the denominator of the given term. We express the numerators as variables and then compute their respective values.

**Complete step-by-step answer:**

To solve the given term by partial fraction method, let us assume

$\dfrac{{23{\text{x}} - 11{{\text{x}}^2}}}{{\left( {{\text{2x - 1}}} \right)\left( {{\text{9 - }}{{\text{x}}^2}} \right)}} = \dfrac{{\text{A}}}{{2{\text{x - 1}}}} + \dfrac{{\text{B}}}{{{\text{3 + x}}}} + \dfrac{{\text{C}}}{{3 - {\text{x}}}}$

Taking LCM on the R.H.S. and then canceling the denominator, we get

$

\therefore {\text{23x - 11}}{{\text{x}}^2} = {\text{A}}\left( {3 + {\text{x}}} \right)\left( {{\text{3 - x}}} \right) + {\text{B}}\left( {2{\text{x - 1}}} \right)\left( {{\text{3 - x}}} \right) + {\text{C}}\left( {2{\text{x - 1}}} \right)\left( {{\text{3 + x}}} \right) \\

\therefore {\text{23x - 11}}{{\text{x}}^2} = {\text{A}}\left( {9 - {{\text{x}}^2}} \right) + {\text{B}}\left( {7{\text{x - 2}}{{\text{x}}^2} - 3} \right) + {\text{C}}\left( {{\text{5x + 2}}{{\text{x}}^2} - 3} \right) \\

\therefore {\text{23x - 11}}{{\text{x}}^2} = {{\text{x}}^2}\left( {{\text{2C - 2B - A}}} \right) + {\text{x}}\left( {7{\text{B + 5C}}} \right) + \left( {{\text{9A - 3B - 3C}}} \right) \\

$

Now, equating the coefficients of like term

2C – 2B – A = -11

7B + 5C = 23

9A – 3B – 3C = 0

Solving above three equations for three variables, we get

A = 1, B = 4, C = -1

Hence, finally we have

$\dfrac{{23{\text{x}} - 11{{\text{x}}^2}}}{{\left( {{\text{2x - 1}}} \right)\left( {{\text{9 - }}{{\text{x}}^2}} \right)}} = \dfrac{1}{{2{\text{x - 1}}}} + \dfrac{4}{{{\text{3 + x}}}} + \dfrac{{ - 1}}{{3 - {\text{x}}}}$

**Note:**In order to solve a term by partial fraction, assuming the given term in terms of some variables like reduces the hectic nature of the problem. The method of splitting into partial fraction becomes very useful while solving integration.

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