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Integrate the rational functions
$\frac{1}{x^2-9}$
$\frac{1}{x^2-9}$
Solution:
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Verified Answer
Let $\frac{1}{x^2-9}=\frac{1}{(x-3)(x+3)} \equiv \frac{A}{x-3}+\frac{B}{x+3}$
$\Rightarrow 1 \equiv \mathrm{A}(x+3)+\mathrm{B}(x-3) \quad \text {...(i) }$
Put $x=3,-3$ in (i), we get : $\mathrm{A}=\frac{1}{6} \& \mathrm{~B}=-\frac{1}{6}$
$\begin{aligned}
&\therefore \quad \int \frac{1}{x^2-9} d x=\frac{1}{6} \int\left[\frac{1}{x-3}-\frac{1}{x+3}\right] d x \\
&=\frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+C
\end{aligned}$
$\Rightarrow 1 \equiv \mathrm{A}(x+3)+\mathrm{B}(x-3) \quad \text {...(i) }$
Put $x=3,-3$ in (i), we get : $\mathrm{A}=\frac{1}{6} \& \mathrm{~B}=-\frac{1}{6}$
$\begin{aligned}
&\therefore \quad \int \frac{1}{x^2-9} d x=\frac{1}{6} \int\left[\frac{1}{x-3}-\frac{1}{x+3}\right] d x \\
&=\frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+C
\end{aligned}$
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