If int frac{1}{(x^{2} + 4)(x^{2} + 9)} dx = A | vedclass

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Question

(A) We use the method of partial fractions to decompose the integrand: $\frac{1}{(x^{2} + 4)(x^{2} + 9)} = \frac{1}{5} \left( \frac{1}{x^{2} + 4} - \f

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$\frac{1}{6}$

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(A) We use the method of partial fractions to decompose the integrand: $\frac{1}{(x^{2} + 4)(x^{2} + 9)} = \frac{1}{5} \left( \frac{1}{x^{2} + 4} - \frac{1}{x^{2} + 9} ight)$.Integrating both sides with respect to $x$:$\int \frac{1}{(x^{2} + 4)(x^{2} + 9)} dx = \frac{1}{5} \int \frac{1}{x^{2} + 2^{2}} dx - \frac{1}{5} \int \frac{1}{x^{2} + 3^{2}} dx$.Using the standard formula $\int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} 	an^{-1} \frac{x}{a} + C$:$= \frac{1}{5} \left( \frac{1}{2} 	an^{-1} \frac{x}{2} ight) - \frac{1}{5} \left( \frac{1}{3} 	an^{-1} \frac{x}{3} ight) + C$.$= \frac{1}{10} 	an^{-1} \frac{x}{2} - \frac{1}{15} 	an^{-1} \frac{x}{3} + C$.Comparing this with $A 	an^{-1} \frac{x}{2} + B 	an^{-1} \frac{x}{3} + C$,we get $A = \frac{1}{10}$ and $B = -\frac{1}{15}$.Therefore,$A - B = \frac{1}{10} - (-\frac{1}{15}) = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$.

If $\int \frac{1}{(x^{2} + 4)(x^{2} + 9)} dx = A \tan^{-1} \frac{x}{2} + B \tan^{-1} \left( \frac{x}{3} \right) + C$,then $A - B =$

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