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Integrate the rational functions
$\frac{5 x}{(x+1)\left(x^2-4\right)}$
$\frac{5 x}{(x+1)\left(x^2-4\right)}$
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Verified Answer
Let $\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{5 x}{(x+1)(x+2)(x-2)}$
$\equiv \frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C}}{x-2} \Rightarrow 5 x \equiv \mathrm{A}(x+2)(x-2)$
$+\mathrm{B}(x+1)(x-2)+\mathrm{C}(x+1)(x+2) \quad$...(i)
Put $\mathrm{x}=-1,-2,2$ in (i) $\Rightarrow \mathrm{A}=\frac{5}{3}, \mathrm{~B}=-\frac{5}{2} \& \mathrm{C}=\frac{5}{6}$
$\begin{aligned}
&\therefore \mathrm{I}=\frac{5}{3} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{x+2}+\frac{5}{6} \int \frac{d x}{x-2} \\
&=\frac{5}{3} \log |x+1|-\frac{5}{2} \log |x+2|+\frac{5}{6} \log |x-2|+\mathrm{C}
\end{aligned}$
$\equiv \frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C}}{x-2} \Rightarrow 5 x \equiv \mathrm{A}(x+2)(x-2)$
$+\mathrm{B}(x+1)(x-2)+\mathrm{C}(x+1)(x+2) \quad$...(i)
Put $\mathrm{x}=-1,-2,2$ in (i) $\Rightarrow \mathrm{A}=\frac{5}{3}, \mathrm{~B}=-\frac{5}{2} \& \mathrm{C}=\frac{5}{6}$
$\begin{aligned}
&\therefore \mathrm{I}=\frac{5}{3} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{x+2}+\frac{5}{6} \int \frac{d x}{x-2} \\
&=\frac{5}{3} \log |x+1|-\frac{5}{2} \log |x+2|+\frac{5}{6} \log |x-2|+\mathrm{C}
\end{aligned}$
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