Tricky integration/functions problem

Question

For $x>0$, let $f(x) = \displaystyle \int_1^x\frac{\ln t}{1+t}dt$. Find the function $f(x) + f(1/x)$ and show that $f(e) + f(1/e) = 1/2$.

Any help would be thoroughly appreciated.

Answer 1

So we have $$f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln(t)}{t \cdot (1+t)} \ dt$$ Adding $$ f(x) + f\left(\frac{1}{x}\right) =\int_{1}^{x} \frac{\ln(t)}{t} \ dt =\frac{(\ln{x})^{2}}{2}$$

Added. \begin{align*} f\left(\frac{1}{x}\right) &= \int_{1}^{1/x} \frac{\ln(t)}{1+t}\ dt \\ &= \int_{1}^{x} \frac{\ln(1/v)}{1+\frac{1}{v}} \cdot -\frac{dv}{v^2} \qquad t\mapsto \frac{1}{v} \\ &=\int_{1}^{x} \frac{\ln(v)}{v \cdot (1+v)} \ dv =\int_{1}^{x} \frac{\ln(t)}{t \cdot (1+t)} \ dt \end{align*}

Answer 2

Let $g(x)=f(x)+f(\frac{1}{x})$. Then $ g^{\prime}(x)=\frac{\ln x}{1+x}+\frac{\ln\frac{1}{x}}{1+\frac{1}{x}}\cdot\left(\frac{-1}{x^2}\right)=\frac{\ln x}{1+x}+\frac{-\ln x}{x+1}\cdot\left(\frac{-x}{x^2}\right)=\frac{x\ln x+\ln x}{x(x+1)}=\frac{(\ln x)(x+1)}{x(x+1)}=\frac{\ln x}{x}$,

so $g(x)=\frac{1}{2}(\ln x)^2+C$, and $C=g(1)=0$.