Let f(x) = int frac{2x}{(x^2+1)(x^2+3)} dx. I | vedclass

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(A) Let $t = x^2$,then $dt = 2x dx$. Substituting this into the integral,we get: $f(x) = \int \frac{dt}{(t+1)(t+3)}$. Using partial fractions: $\frac{

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$\frac{1}{2}(\log_e 17 - \log_e 19)$

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(A) Let $t = x^2$,then $dt = 2x dx$. Substituting this into the integral,we get: $f(x) = \int \frac{dt}{(t+1)(t+3)}$. Using partial fractions: $\frac{1}{(t+1)(t+3)} = \frac{1}{2} \left( \frac{1}{t+1} - \frac{1}{t+3} ight)$. Integrating both sides: $f(x) = \frac{1}{2} \int \left( \frac{1}{t+1} - \frac{1}{t+3} ight) dt = \frac{1}{2} (\ln|t+1| - \ln|t+3|) + C$. Substituting $t = x^2$ back: $f(x) = \frac{1}{2} \ln \left( \frac{x^2+1}{x^2+3} ight) + C$. Given $f(3) = \frac{1}{2}(\ln 10 - \ln 12) + C = \frac{1}{2}(\ln 5 - \ln 6) + C$. Since $\frac{1}{2}(\ln 10 - \ln 12) = \frac{1}{2}(\ln(2 	imes 5) - \ln(2 	imes 6)) = \frac{1}{2}(\ln 5 - \ln 6)$,we find $C = 0$. Thus,$f(x) = \frac{1}{2} \ln \left( \frac{x^2+1}{x^2+3} ight)$. Calculating $f(4)$: $f(4) = \frac{1}{2} \ln \left( \frac{4^2+1}{4^2+3} ight) = \frac{1}{2} \ln \left( \frac{17}{19} ight) = \frac{1}{2} (\ln 17 - \ln 19)$.

Let $f(x) = \int \frac{2x}{(x^2+1)(x^2+3)} dx$. If $f(3) = \frac{1}{2}(\log_e 5 - \log_e 6)$,then $f(4)$ is equal to

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