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

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(A) Given $f(x) = \int \frac{x}{(x^2+1)(x^2+3)} dx$. Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{1}{2} dt$. Substituting these into the integral,

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$\frac{1}{4} \log \left(\frac{1}{3}\right)$

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(A) Given $f(x) = \int \frac{x}{(x^2+1)(x^2+3)} dx$. Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{1}{2} dt$. Substituting these into the integral,we get: $f(x) = \frac{1}{2} \int \frac{1}{(t+1)(t+3)} dt$. Using partial fractions: $\frac{1}{(t+1)(t+3)} = \frac{1}{2} \left( \frac{1}{t+1} - \frac{1}{t+3} \right)$. Thus,$f(x) = \frac{1}{2} \cdot \frac{1}{2} \int \left( \frac{1}{t+1} - \frac{1}{t+3} \right) dt = \frac{1}{4} [\log|t+1| - \log|t+3|] + C = \frac{1}{4} \log \left( \frac{x^2+1}{x^2+3} \right) + C$. Given $f(3) = \frac{1}{4} \log \left( \frac{5}{6} \right)$,we have: $\frac{1}{4} \log \left( \frac{3^2+1}{3^2+3} \right) + C = \frac{1}{4} \log \left( \frac{10}{12} \right) + C = \frac{1}{4} \log \left( \frac{5}{6} \right) + C$. Comparing this with the given value,we find $C = 0$. Therefore,$f(x) = \frac{1}{4} \log \left( \frac{x^2+1}{x^2+3} \right)$. Finally,$f(0) = \frac{1}{4} \log \left( \frac{0^2+1}{0^2+3} \right) = \frac{1}{4} \log \left( \frac{1}{3} \right)$.

Let $f(x) = \int \frac{x}{(x^2+1)(x^2+3)} dx$. If $f(3) = \frac{1}{4} \log \left(\frac{5}{6}\right)$,then find $f(0)$.

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