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

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Question

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

Vedclass · Accepted Answer

$\frac{{11}}{{12}}, \frac{5}{6}$

Vedclass · Answer

(A) Let $I = \int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} - 4)}}} dx$. Using partial fractions,let $\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} - 4)}} = \frac{A}{{{x^2} - 1}} + \frac{B}{{{x^2} - 4}}$. Then $2{x^2} + 3 = A({x^2} - 4) + B({x^2} - 1)$. Comparing coefficients of ${x^2}$ and constant terms: $A + B = 2$ $(i)$ $-4A - B = 3$ $(ii)$ Adding $(i)$ and $(ii)$,$-3A = 5 \implies A = -\frac{5}{3}$. Substituting $A$ in $(i)$,$B = 2 + \frac{5}{3} = \frac{11}{3}$. So,$I = \int {\frac{{ - 5/3}}{{{x^2} - 1}} dx} + \int {\frac{{11/3}}{{{x^2} - 4}} dx}$. Using $\int {\frac{{dx}}{{{x^2} - {a^2}}}} = \frac{1}{{2a}}\log \left| {\frac{{x - a}}{{x + a}}} \right| + c$: $I = -\frac{5}{3} \cdot \frac{1}{2} \log \left| {\frac{{x - 1}}{{x + 1}}} \right| + \frac{{11}}{3} \cdot \frac{1}{4} \log \left| {\frac{{x - 2}}{{x + 2}}} \right| + c$. $I = -\frac{5}{6} \log \left| {\frac{{x - 1}}{{x + 1}}} \right| + \frac{{11}}{{12}} \log \left| {\frac{{x - 2}}{{x + 2}}} \right| + c$. $I = \frac{5}{6} \log \left| {\frac{{x + 1}}{{x - 1}}} \right| + \frac{{11}}{{12}} \log \left| {\frac{{x - 2}}{{x + 2}}} \right| + c$. Comparing with $\log {\left( {\frac{{x - 2}}{{x + 2}}} \right)^a}{\left( {\frac{{x + 1}}{{x - 1}}} \right)^b} + c$,we get $a = \frac{{11}}{{12}}$ and $b = \frac{5}{6}$.

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

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