If 729 int_1^3 frac{1}{x^3(x^2+9)^2} dx = a + | vedclass

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(A) Let $I = 729 \int_1^3 \frac{1}{x^3(x^2+9)^2} dx$.Using partial fractions,we write $\frac{1}{x^3(x^2+9)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}

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$4$

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(A) Let $I = 729 \int_1^3 \frac{1}{x^3(x^2+9)^2} dx$.Using partial fractions,we write $\frac{1}{x^3(x^2+9)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx+E}{x^2+9} + \frac{Fx+G}{(x^2+9)^2}$.Solving for the coefficients,we get $A = -\frac{2}{729}$,$B = 0$,$C = \frac{1}{81}$,$D = \frac{2}{729}$,$E = 0$,$F = \frac{1}{81}$,$G = 0$.Substituting these values into the integral:$I = 729 \int_1^3 \left( -\frac{2}{729x} + \frac{1}{81x^3} + \frac{2x}{729(x^2+9)} + \frac{x}{81(x^2+9)^2} \right) dx$$I = \int_1^3 \left( -\frac{2}{x} + \frac{9}{x^3} + \frac{2x}{x^2+9} + \frac{9x}{(x^2+9)^2} \right) dx$Integrating term by term:$I = \left[ -2 \log|x| - \frac{9}{2x^2} + \log(x^2+9) - \frac{9}{2(x^2+9)} \right]_1^3$$I = \left[ \log\left(\frac{x^2+9}{x^2}\right) - \frac{9}{2x^2} - \frac{9}{2(x^2+9)} \right]_1^3$Evaluating at the limits:$I = \left( \log\left(\frac{18}{9}\right) - \frac{9}{18} - \frac{9}{36} \right) - \left( \log\left(\frac{10}{1}\right) - \frac{9}{2} - \frac{9}{20} \right)$$I = \log 2 - \frac{1}{2} - \frac{1}{4} - \log 10 + \frac{9}{2} + \frac{9}{20}$$I = \log\left(\frac{2}{10}\right) + \left( 4 - \frac{1}{4} + \frac{9}{20} \right) = \log\left(\frac{1}{5}\right) + \left( \frac{80 - 5 + 9}{20} \right) = \log\left(\frac{1}{5}\right) + \frac{84}{20} = \frac{21}{5} + \log\left(\frac{1}{5}\right)$Comparing with $a + \log b$,we get $a = \frac{21}{5}$ and $b = \frac{1}{5}$.Therefore,$a - b = \frac{21}{5} - \frac{1}{5} = \frac{20}{5} = 4$.

If $729 \int_1^3 \frac{1}{x^3(x^2+9)^2} dx = a + \log b$,then $a - b =$

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