1. Integrate the given function

We need

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

Let

$$$$

Then

$$f(x)=\int \frac{du}{(u+1)(u+3)}.$$

Now use partial fractions:

$$\frac{1}{(u+1)(u+3)}=\frac{A}{u+1}+\frac{B}{u+3}.$$

So,

$$1=A(u+3)+B(u+1).$$

Putting $u=-1$ gives

$$$$

and putting $u=-3$ gives

$$$$

Hence,

$$\frac{1}{(u+1)(u+3)}=\frac12\left(\frac{1}{u+1}-\frac{1}{u+3}\right).$$

Therefore,

$$f(x)=\frac12\int \left(\frac{1}{u+1}-\frac{1}{u+3}\right)du =\frac12\big(\ln(u+1)-\ln(u+3)\big)+C.$$

Substituting $u=x^2$,

$$f(x)=\frac12\left(\ln(x^2+1)-\ln(x^2+3)\right)+C.$$

2. Use the condition $f(3)=\frac12(\ln 5-\ln 6)$ to find $C$

At $x=3$,

$$f(3)=\frac12\left(\ln(10)-\ln(12)\right)+C =\frac12\ln\left(\frac{10}{12}\right)+C =\frac12\ln\left(\frac56\right)+C.$$

Given,

$$f(3)=\frac12(\ln 5-\ln 6)=\frac12\ln\left(\frac56\right).$$

Thus,

$$$$

So,

$$f(x)=\frac12\left(\ln(x^2+1)-\ln(x^2+3)\right).$$

3. Find $f(4)$

$$f(4)=\frac12\left(\ln(16+1)-\ln(16+3)\right) =\frac12(\ln 17-\ln 19).$$

Thus,

$$f(4)=\frac12(\ln 17-\ln 19).$$

4. Match with options

This is exactly Option D.