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      <text>9.3.12 (Partial franctions - repeated linear factors)</text>
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      <text><![CDATA[<p>One other possibility that we could have, when trying to use partial fractions is where we have a repeated linear factor e.g. an expression like \(\dfrac{2x+1}{(x-1)^2}\) . Recall that the aim of the partial fractions decomposition is to arrive at a sum of terms where the numerators are constants. We can achieve this using terms with different powers of \((x-1)\) in the denominator:
\[
&nbsp;\dfrac{2x+1}{(x-1)^2} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2} = \dfrac{A(x-1)+B}{(x-1)^2}
\]
From this, we can easily find \(A\) and \(B\) so that \(A(x-1)+B=2x+1\) -- from the coefficient of \(x\) we see \(A=2\) and then looking at the constant term, we get \(B=3\).
This technique works more generally. For proper rational functions with a <i>repeated linear factor</i> in the denominator, the partial fractions decomposition consists of terms whose denominators are powers of the linear factor. In particular:
\[ 
\dfrac{p(x)}{(x-a)^n} = \dfrac{A_1}{x-a}+\dfrac{A_2}{(x-a)^2}+\cdots+\dfrac{A_n}{(x-a)^n} 
\]
So for instance:</p>
<p></p>
<ul>
<li>if the denominator has \((x-a)^2\) then we add terms \(\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}\).<br></li>
<li>if the denominator has \((x-a)^3\) then we add terms \(\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{(x-a)^3}\).<br></li>
</ul>]]></text>
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