If int frac{2x^2+3}{(x^2-1)(x^2-4)} dx = log | vedclass

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

Vedclass · Accepted Answer

$\frac{21}{12}$

Vedclass · Answer

(B) Let $I = \int \frac{2x^2+3}{(x^2-1)(x^2-4)} dx$. 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 $t=1$,$5 = A(-3) \implies A = -\frac{5}{3}$. For $t=4$,$11 = B(3) \implies B = \frac{11}{3}$. So,$\frac{2x^2+3}{(x^2-1)(x^2-4)} = \frac{-5/3}{x^2-1} + \frac{11/3}{x^2-4}$. Integrating,$I = -\frac{5}{3} \int \frac{dx}{x^2-1} + \frac{11}{3} \int \frac{dx}{x^2-4}$. $I = -\frac{5}{3} \cdot \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| + \frac{11}{3} \cdot \frac{1}{4} \log \left| \frac{x-2}{x+2} \right| + c$. $I = \frac{5}{6} \log \left| \frac{x+1}{x-1} \right| + \frac{11}{12} \log \left| \frac{x-2}{x+2} \right| + c$. Comparing with the given form,$a = \frac{11}{12}$ and $b = \frac{5}{6} = \frac{10}{12}$. Thus,$a+b = \frac{11}{12} + \frac{10}{12} = \frac{21}{12}$.

If $\int \frac{2x^2+3}{(x^2-1)(x^2-4)} dx = \log \left[\left(\frac{x-2}{x+2}\right)^a \cdot \left(\frac{x+1}{x-1}\right)^b\right] + c$,(where $c$ is the constant of integration),then the value of $a+b$ is equal to

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