Generalise it first

Define $\psi(x)=f(x)+f(1/x)$ Differentiating both sides w.r.t. $x$ using generalized Leibniz integral rule , we have $\psi'(x)=\frac{\ln x}{1+x} + \frac{\ln(1/x)}{1+1/x}. -\frac{1}{x^2}=\frac{\ln(x)}{x}$ Hence, $\psi(x)= \int \frac{\ln(x)}{x} dx = \frac{1}{2}\ln^2(x)+C$ where $C$ is the constant of integration. Putting $x=1$ , we see that the lower and upper limits of both the integrals coincide (1). Hence $\psi(1)=0$ . This in turn implies that $C=0$ and hence $\psi(x)= \frac{1}{2}\ln^2(x)$ We require $f(e)+f(1/e)=\psi(e)=\frac{1}{2}\hspace{5pt} \blacksquare$

$f(e)+f(1/e)=\int\limits_1^e \frac{lnt}{t+1}dt + \int\limits_{1}^{1/e} \frac{lnt}{t+1}dt$

put $t=1/u$ in the second integral and you will get,

$=\int\limits_1^e \frac{lnt}{t+1}dt + \int\limits_1^e \frac{lnu}{u(u+1)}du$ $=\int\limits_1^e \frac{lnt}{t}dt =1/2$

using $\int\limits_a^b f(x)dx=\int\limits_a^b f(t)dt$

Use substitution for t=1/z.