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Integrate the function
$\int \frac{5 x}{(x+1)\left(x^2+9\right)} d x=I($ say $)$
$\int \frac{5 x}{(x+1)\left(x^2+9\right)} d x=I($ say $)$
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Let $\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+9}$ $\Rightarrow 5 \mathrm{x}=\mathrm{A}\left(\mathrm{x}^2+9\right)+(\mathrm{Bx}+\mathrm{c})(\mathrm{x}+1) \ldots$ (i)
Putting $x=-1$ in (i) we get:, $A=\frac{-1}{2}$
Comparing coefficients: $\mathrm{B}=\frac{1}{2}$ and $\mathrm{C}=\frac{9}{2}$
$I=\frac{-1}{2} \int \frac{1}{x+1} d x+\frac{1}{4} \int \frac{2 x}{x^2+9} d x+\frac{9}{2} \int \frac{1}{x^2+9} d x$
$=-\frac{1}{2} \log |x+1|+\frac{1}{4} \log \left|x^2+9\right|+\frac{3}{2} \tan ^{-1}\left(\frac{x}{3}\right)+c$
Putting $x=-1$ in (i) we get:, $A=\frac{-1}{2}$
Comparing coefficients: $\mathrm{B}=\frac{1}{2}$ and $\mathrm{C}=\frac{9}{2}$
$I=\frac{-1}{2} \int \frac{1}{x+1} d x+\frac{1}{4} \int \frac{2 x}{x^2+9} d x+\frac{9}{2} \int \frac{1}{x^2+9} d x$
$=-\frac{1}{2} \log |x+1|+\frac{1}{4} \log \left|x^2+9\right|+\frac{3}{2} \tan ^{-1}\left(\frac{x}{3}\right)+c$
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