If int frac{2 x^{2}+3}{left(x^{2}-1right)left | vedclass

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Question

(A) Let $I = \int \frac{2 x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}+4\right)} d x$. Using partial fractions,let $x^2 = t$. Then $\frac{2t+3}{(t-1)(t+4)

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$a=\frac{1}{2}, \quad b=\frac{1}{2}$

Vedclass · Answer

(A) Let $I = \int \frac{2 x^{2}+3}{\left(x^{2}-1ight)\left(x^{2}+4ight)} d x$. Using partial fractions,let $x^2 = t$. Then $\frac{2t+3}{(t-1)(t+4)} = \frac{A}{t-1} + \frac{B}{t+4}$. $2t+3 = A(t+4) + B(t-1)$. For $t=1$,$5 = 5A \implies A=1$. For $t=-4$,$-5 = -5B \implies B=1$. Thus,$\frac{2x^2+3}{(x^2-1)(x^2+4)} = \frac{1}{x^2-1} + \frac{1}{x^2+4}$. Integrating both sides,$I = \int \frac{1}{x^2-1} dx + \int \frac{1}{x^2+2^2} dx$. $I = \frac{1}{2(1)} \log \left| \frac{x-1}{x+1} ight| + \frac{1}{2} 	an^{-1} \left( \frac{x}{2} ight) + C$. Comparing this with the given expression $a \log \left| \frac{x-1}{x+1} ight| + b 	an^{-1} \left( \frac{x}{2} ight) + C$,we get $a = \frac{1}{2}$ and $b = \frac{1}{2}$.

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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