If int frac{2 x^2-3}{left(x^2-4right)left(x^2 | vedclass

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

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Let $I = \int \frac{2 x^2-3}{\left(x^2-4ight)\left(x^2+1ight)} d x$. Using partial fractions,let $\frac{2 x^2-3}{\left(x^2-4ight)\left(x^2+1ight)} = \frac{P}{x^2-4} + \frac{Q}{x^2+1}$. $2x^2 - 3 = P(x^2+1) + Q(x^2-4) = (P+Q)x^2 + (P-4Q)$. Comparing coefficients: $P+Q = 2$ and $P-4Q = -3$. Subtracting the equations: $5Q = 5 \implies Q = 1$. Then $P = 1$. So,$I = \int \frac{1}{x^2-4} d x + \int \frac{1}{x^2+1} d x$. $I = \frac{1}{2(2)} \log \left| \frac{x-2}{x+2} ight| + 	an^{-1} x + K$. $I = 	an^{-1} x + \frac{1}{4} \log (x-2) - \frac{1}{4} \log (x+2) + K$. Comparing with $A 	an^{-1} x + B \log (x-2) + C \log (x+2)$,we get $A = 1$,$B = \frac{1}{4}$,$C = -\frac{1}{4}$. Then $6A + 7B - 5C = 6(1) + 7(\frac{1}{4}) - 5(-\frac{1}{4}) = 6 + \frac{7}{4} + \frac{5}{4} = 6 + \frac{12}{4} = 6 + 3 = 9$.

If $\int \frac{2 x^2-3}{\left(x^2-4\right)\left(x^2+1\right)} d x=A \tan^{-1} x+B \log (x-2)+C \log (x+2)$,then $6 A+7 B-5 C=$

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