1. We are given

$$f(x)=\int_1^x \frac{\ln t}{1+t}\,dt, \qquad x>0.$$

We need to find

$$f(e)+f\left(\frac1e\right).$$

2. Write both terms using the definition:

$$f(e)=\int_1^e \frac{\ln t}{1+t}\,dt,$$

and

$$f\left(\frac1e\right)=\int_1^{1/e} \frac{\ln t}{1+t}\,dt.$$

So,

$$f(e)+f\left(\frac1e\right)=\int_1^e \frac{\ln t}{1+t}\,dt+\int_1^{1/e} \frac{\ln t}{1+t}\,dt.$$

3. Reverse the limits in the second integral:

$$\int_1^{1/e} \frac{\ln t}{1+t}\,dt=-\int_{1/e}^1 \frac{\ln t}{1+t}\,dt.$$

Hence

$$f(e)+f\left(\frac1e\right)=\int_1^e \frac{\ln t}{1+t}\,dt-\int_{1/e}^1 \frac{\ln t}{1+t}\,dt.$$

4. In the second integral, use the substitution

$$t=\frac1u \quad \Rightarrow \quad dt=-\frac1{u^2}du.$$

When $t=\frac1e$, $u=e$, and when $t=1$, $u=1$.

Then

$$\int_{1/e}^1 \frac{\ln t}{1+t}\,dt =\int_e^1 \frac{\ln(1/u)}{1+1/u}\left(-\frac1{u^2}\right)du.$$

Now simplify:

$$\ln(1/u)=-\ln u,$$

and

$$1+\frac1u=\frac{u+1}{u}.$$

Therefore,

$$\frac{\ln(1/u)}{1+1/u}\left(-\frac1{u^2}\right) =\frac{-\ln u}{(u+1)/u}\left(-\frac1{u^2}\right) =\frac{\ln u}{u(u+1)}.$$

So,

$$\int_{1/e}^1 \frac{\ln t}{1+t}\,dt=\int_e^1 \frac{\ln u}{u(u+1)}du=-\int_1^e \frac{\ln u}{u(u+1)}du.$$

Thus,

$$f(e)+f\left(\frac1e\right)=\int_1^e \frac{\ln t}{1+t}dt+\int_1^e \frac{\ln t}{t(1+t)}dt.$$

5. Combine the integrands:

$$\frac{\ln t}{1+t}+\frac{\ln t}{t(1+t)} =\ln t\left(\frac1{1+t}+\frac1{t(1+t)}\right).$$

Take LCM:

$$\frac1{1+t}+\frac1{t(1+t)} =\frac{t+1}{t(1+t)}=\frac1t.$$

Hence,

$$f(e)+f\left(\frac1e\right)=\int_1^e \frac{\ln t}{t}\,dt.$$

6. Now evaluate:

$$\int \frac{\ln t}{t}\,dt=\frac{(\ln t)^2}{2}.$$

Therefore,

$$\int_1^e \frac{\ln t}{t}\,dt=\left[\frac{(\ln t)^2}{2}\right]_1^e =\frac{(\ln e)^2}{2}-\frac{(\ln 1)^2}{2} =\frac{1^2}{2}-0=\frac12.$$

7. Final answer:

$$f(e)+f\left(\frac1e\right)=\frac12.$$

So the correct option is A.