Let $f(x) =\int^x_1 \frac{\ln t}{1+t}dt ;$ for $x >0$ then find the value of $f(e) +f(\frac{1}{e})$ Problem : 
Let $f(x) =\int^x_1 \frac{\ln t}{1+t}dt ;$ for $x >0$ then find the value of $f(e) +f(\frac{1}{e})$
please prove some hint on this as I am not getting any clue how to proceed here, thanks 
 A: Sub $t = 1/u$ in the integral; then
$$f(x) = \int_1^{1/x} \frac{du}{u} \frac{\log{u}}{1+u} = \int_1^{1/x} du \, \log{u} \, \left (\frac1u - \frac1{1+u} \right ) = \frac12 \log^2{x} - f \left (\frac1x \right )$$
Thus
$$f(e) + f \left (\frac1e \right ) = \frac12 $$
A: Given $$f(x) = \int_{1}^{x}\frac{\ln t}{1+t}dt$$ and $$f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}}\frac{\ln t}{1+t}dt$$
So $$F(x) = \int_{1}^{x}\frac{\ln t}{1+t}dt+\int_{1}^{\frac{1}{x}}\frac{\ln t}{1+t}dt$$
Now Using Differentiation under Integral Sign, We get
$$F'(x)=\frac{\ln x}{1+x}+\frac{\ln x}{(1+x)x} = \frac{\ln x\cdot (1+x)}{(1+x)\cdot x}$$
So we get $$F'(x) = \frac{\ln x}{x} \Rightarrow \int F'(x)dx = \int \frac{\ln x}{x}dx$$
So we get $$F(x) = \frac{(\ln x)^2}{2}+\mathcal{C}$$
Now at $x=1\;,$ We get $$F(1) = f(1)+f\left(\frac{1}{1}\right)=0$$
So we get $$F(1)= \frac{(\ln(1))^2}{2}+\mathcal{C}$$
So we get $\mathcal{C=0}\;,$ Then $$F(x)=f(x)+f\left(\frac{1}{x}\right)=\frac{(\ln x)^2}{2}$$
So we get $$F(e)=f(e)+f\left(\frac{1}{e}\right)=\frac{(\ln e)^2}{2}=\frac{1}{2}$$
