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Integrate the rational functions
$\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}$
$\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}$
Solution:
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Verified Answer
Put $x^2=y, I=1-\frac{2(2 y+5)}{(y+3)(y+4)} \quad \ldots(i)$
Let $; \frac{2 y+5}{(y+3)(y+4)}=\frac{A}{y+3}+\frac{B}{y+4}$,
$2 \mathrm{y}+5 \equiv \mathrm{A}(\mathrm{y}+4)+\mathrm{B}(\mathrm{y}+3)$
Put $\mathrm{y}=-3, \therefore \mathrm{A}=-1$, Put $\mathrm{y}=-4, \therefore \mathrm{B}=3$
$\begin{aligned}
&\therefore I=\int d x+2 \int \frac{d x}{x^2+3}+6 \int \frac{d x}{x^2+4} \\
&=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+c
\end{aligned}$
Let $; \frac{2 y+5}{(y+3)(y+4)}=\frac{A}{y+3}+\frac{B}{y+4}$,
$2 \mathrm{y}+5 \equiv \mathrm{A}(\mathrm{y}+4)+\mathrm{B}(\mathrm{y}+3)$
Put $\mathrm{y}=-3, \therefore \mathrm{A}=-1$, Put $\mathrm{y}=-4, \therefore \mathrm{B}=3$
$\begin{aligned}
&\therefore I=\int d x+2 \int \frac{d x}{x^2+3}+6 \int \frac{d x}{x^2+4} \\
&=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+c
\end{aligned}$
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