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If $\int \frac{2 x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}+4\right)} d x=a \log \left|\frac{x-1}{x+1}\right|+b \tan ^{-1}\left(\frac{x}{2}\right)+C$, then
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The correct answer is:
$\mathrm{a}=\frac{1}{2}, \quad \mathrm{~b}=\frac{1}{2}$
$$
\text { Let } \begin{aligned}
I &=\int \frac{2 x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}+4\right)} d x=\int\left[\frac{1}{x^{2}-1}+\frac{1}{x^{2}+4}\right] d x \\
I &=\frac{1}{2(1)} \log \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \tan ^{-1} \frac{x}{2}+C
\end{aligned}
$$
Comparing with given data, we get $a=\frac{1}{2}, b=\frac{1}{2}$
\text { Let } \begin{aligned}
I &=\int \frac{2 x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}+4\right)} d x=\int\left[\frac{1}{x^{2}-1}+\frac{1}{x^{2}+4}\right] d x \\
I &=\frac{1}{2(1)} \log \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \tan ^{-1} \frac{x}{2}+C
\end{aligned}
$$
Comparing with given data, we get $a=\frac{1}{2}, b=\frac{1}{2}$
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