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

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Question

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

Vedclass · Accepted Answer

$(\frac{1}{2}, \frac{1}{2})$

Vedclass · Answer

(D) Let $I = \int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} + 4)}}dx}$.Using partial fractions,let $t = x^2$. 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 $t=1$,$5 = 5A \Rightarrow A=1$.For $t=-4$,$-5 = -5B \Rightarrow B=1$.Thus,$\frac{2{x^2} + 3}{{({x^2} - 1)({x^2} + 4)}} = \frac{1}{{({x^2} - 1)}} + \frac{1}{{({x^2} + 4)}}$.$I = \int {\frac{{dx}}{{({x^2} - 1)}} + \int {\frac{{dx}}{{{x^2} + 2^2}}} }$.Using standard integrals $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log|\frac{x-a}{x+a}|$ and $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})$:$I = \frac{1}{2(1)} \log \left| \frac{x-1}{x+1} ight| + \frac{1}{2} 	an^{-1}(\frac{x}{2}) + c$.Comparing with $a \log \left( \frac{x-1}{x+1} ight) + b 	an^{-1}(\frac{x}{2}) + c$,we get $a = \frac{1}{2}$ and $b = \frac{1}{2}$.

If $\int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} + 4)}}dx = a\log \left( {\frac{{x - 1}}{{x + 1}}} \right) + b\tan ^{ - 1}\frac{x}{2} + c}$,then the values of $a$ and $b$ are:

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