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

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(D) Let $t = x^2$, then $dt = 2x dx$, so $x dx = \frac{dt}{2}$.$f(x) = \int \frac{1}{2(t+4)(t+9)} dt$.Using partial fractions: $\frac{1}{(t+4)(t+9)} =

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$-\frac{1}{10} \log(2)$

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(D) Let $t = x^2$, then $dt = 2x dx$, so $x dx = \frac{dt}{2}$.$f(x) = \int \frac{1}{2(t+4)(t+9)} dt$.Using partial fractions: $\frac{1}{(t+4)(t+9)} = \frac{1}{5} (\frac{1}{t+4} - \frac{1}{t+9})$.$f(x) = \frac{1}{10} \int (\frac{1}{x^2+4} - \frac{1}{x^2+9}) dx$.$f(x) = \frac{1}{10} [\frac{1}{2} 	an^{-1}(\frac{x}{2}) - \frac{1}{3} 	an^{-1}(\frac{x}{3})] + C$.Wait, the integral is $\int \frac{x}{(x^2+4)(x^2+9)} dx = \frac{1}{2} \int \frac{dt}{(t+4)(t+9)} = \frac{1}{10} \ln|\frac{t+4}{t+9}| + C = \frac{1}{10} \ln|\frac{x^2+4}{x^2+9}| + C$.Given $f(0) = \frac{1}{10} \ln(\frac{4}{9}) + C = \frac{1}{10} \ln((\frac{2}{3})^2) + C = \frac{1}{5} \ln(\frac{2}{3}) + C$.Since $f(0) = \frac{1}{5} \ln(\frac{2}{3})$, we get $C = 0$.Thus, $f(x) = \frac{1}{10} \ln(\frac{x^2+4}{x^2+9})$.$f(1) = \frac{1}{10} \ln(\frac{1+4}{1+9}) = \frac{1}{10} \ln(\frac{5}{10}) = \frac{1}{10} \ln(\frac{1}{2}) = -\frac{1}{10} \ln(2)$.

If $f(x) = \int \frac{x}{(x^2+4)(x^2+9)} dx$ and $f(0) = \frac{1}{5} \log(\frac{2}{3})$, then $f(1) = $

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