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

We are given $$f(x) = \int \frac{2x}{(x^2+1)(x^2+3)}\,dx$$ along with the condition $$f(3) = \frac{1}{2}(\ln 5 - \ln 6)\,. $$ Substituting $$u = x^2$$ so that $$du = 2x\,dx$$ transforms the integral into a function of $$u$$, and integration by partial fractions yields:

$$ f(x) = \int \frac{du}{(u+1)(u+3)} = \frac{1}{2}\int\Bigl(\frac{1}{u+1}-\frac{1}{u+3}\Bigr)\,du = \frac{1}{2}\ln\frac{u+1}{u+3} + C = \frac{1}{2}\ln\frac{x^2+1}{x^2+3} + C. $$

Next, imposing the condition at $$x=3$$ gives

$$ f(3) = \frac{1}{2}\ln\frac{10}{12} + C = \frac{1}{2}\ln\frac{5}{6} + C = \frac{1}{2}(\ln 5 - \ln 6) + C. $$

Since this must equal $$\frac{1}{2}(\ln 5 - \ln 6)$$, it follows that $$C=0$$. Therefore,

$$ f(x) = \frac{1}{2}\ln\frac{x^2+1}{x^2+3}. $$

Finally, evaluating at $$x=4$$ yields

$$ f(4) = \frac{1}{2}\ln\frac{17}{19} = \frac{1}{2}(\ln 17 - \ln 19). $$

Therefore, the value of interest is $$\frac{1}{2}(\ln 17 - \ln 19)\,. $$