If int {frac{{(2{x^2} + 1),dx}}{{({x^2} - 4)( | vedclass

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

Vedclass · Accepted Answer

$1/2, 3/4$

Vedclass · Answer

(A) Let $I = \int {\frac{{2{x^2} + 1}}{{({x^2} - 4)({x^2} - 1)}}dx}$.Using partial fractions,we write $\frac{{2{x^2} + 1}}{{({x^2} - 4)({x^2} - 1)}} = \frac{A}{{x^2 - 4}} + \frac{B}{{x^2 - 1}}$.$2{x^2} + 1 = A(x^2 - 1) + B(x^2 - 4)$.For $x^2 = 1$,$3 = B(1 - 4) \implies 3 = -3B \implies B = -1$.For $x^2 = 4$,$9 = A(4 - 1) \implies 9 = 3A \implies A = 3$.So,$\frac{{2{x^2} + 1}}{{({x^2} - 4)({x^2} - 1)}} = \frac{3}{{x^2 - 4}} - \frac{1}{{x^2 - 1}}$.$I = \int {\frac{3}{{x^2 - 2^2}}dx} - \int {\frac{1}{{x^2 - 1^2}}dx}$.Using the formula $\int {\frac{1}{{x^2 - a^2}}dx = \frac{1}{{2a}}\log \left| {\frac{{x - a}}{{x + a}}} \right| + C}$,we get:$I = 3 \cdot \frac{1}{{2(2)}}\log \left| {\frac{{x - 2}}{{x + 2}}} \right| - \frac{1}{{2(1)}}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| + C$.$I = \frac{3}{4}\log \left| {\frac{{x - 2}}{{x + 2}}} \right| + \frac{1}{2}\log \left| {\frac{{x + 1}}{{x - 1}}} \right| + C$.Comparing this with $\log \left[ {{{\left( {\frac{{x + 1}}{{x - 1}}} \right)}^a}{{\left( {\frac{{x - 2}}{{x + 2}}} \right)}^b}} \right] + C$,we get $a = 1/2$ and $b = 3/4$.

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

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