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Integrate the function
$\int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)} d x$
$\int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)} d x$
Solution:
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Verified Answer
$\frac{1}{(\mathrm{y}+1)(\mathrm{y}+4)},=\frac{\mathrm{A}}{\mathrm{y}+1}+\frac{\mathrm{B}}{\mathrm{y}+4}$ where $\mathrm{x}^2=\mathrm{y}$
$\Rightarrow 1=\mathrm{A}(\mathrm{y}+4)+\mathrm{B}(\mathrm{y}+1) \quad \ldots(i)$
Put $\mathrm{y}=-1,-4$ in (i) $\Rightarrow \mathrm{A}=\frac{1}{3}$ and $\mathrm{B}=-\frac{1}{3}$
$\therefore \mathrm{I}=\frac{1}{3} \int \frac{\mathrm{dx}}{\mathrm{x}^2+1}-\frac{1}{3} \int \frac{\mathrm{dx}}{\mathrm{x}^2+4}$
$=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+c$
$\Rightarrow 1=\mathrm{A}(\mathrm{y}+4)+\mathrm{B}(\mathrm{y}+1) \quad \ldots(i)$
Put $\mathrm{y}=-1,-4$ in (i) $\Rightarrow \mathrm{A}=\frac{1}{3}$ and $\mathrm{B}=-\frac{1}{3}$
$\therefore \mathrm{I}=\frac{1}{3} \int \frac{\mathrm{dx}}{\mathrm{x}^2+1}-\frac{1}{3} \int \frac{\mathrm{dx}}{\mathrm{x}^2+4}$
$=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+c$
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