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Integrate the rational functions
$\frac{2}{(1-x)\left(1+x^2\right)}$
$\frac{2}{(1-x)\left(1+x^2\right)}$
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Verified Answer
Let $\frac{2}{(1-x)\left(1+x^2\right)}=\frac{\mathrm{A}}{1-x}+\frac{\mathrm{B} x+\mathrm{C}}{1+x^2}$
$\Rightarrow 2=\mathrm{A}\left(1+x^2\right)+(\mathrm{B} x+\mathrm{C})(1-x) \quad \ldots(i)$
Putting $x=1$ in (i), we get; $A=1$
Also $0=\mathrm{A}-\mathrm{B}$ and $2=\mathrm{A}+\mathrm{C} \Rightarrow \mathrm{B}=\mathrm{A}=1 \& \mathrm{C}=1$
$\begin{aligned}
&\therefore \mathrm{I}=\int \frac{1}{1-x} d x+\int \frac{x}{1+x^2} d x+\int \frac{1}{1+x^2} d x \\
&=-\log |1-x|+\frac{1}{2} \log \left|1+x^2\right|+\tan ^{-1} x+\mathrm{C}
\end{aligned}$
$\Rightarrow 2=\mathrm{A}\left(1+x^2\right)+(\mathrm{B} x+\mathrm{C})(1-x) \quad \ldots(i)$
Putting $x=1$ in (i), we get; $A=1$
Also $0=\mathrm{A}-\mathrm{B}$ and $2=\mathrm{A}+\mathrm{C} \Rightarrow \mathrm{B}=\mathrm{A}=1 \& \mathrm{C}=1$
$\begin{aligned}
&\therefore \mathrm{I}=\int \frac{1}{1-x} d x+\int \frac{x}{1+x^2} d x+\int \frac{1}{1+x^2} d x \\
&=-\log |1-x|+\frac{1}{2} \log \left|1+x^2\right|+\tan ^{-1} x+\mathrm{C}
\end{aligned}$
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