For $x0$, let $f(x)=\int_{1}^{x} \frac{\ln t}{1+t} d t$, $f(e)+f\left(\frac{1}{e}\right)=\frac{1}{

Question

For $x>0$, let $f(x)=\int_{1}^{x} \frac{\ln t}{1+t} d t$,  $f(e)+f\left(\frac{1}{e}\right)=\frac{1}{n}$.

Professor PC Thomas Classes · Answer

$f(x)=\int_{1}^{x} \frac{\ln (t)}{1+t} d t$\ $f\left(\frac{1}{x}ight)=\int_{1}^{1 / x} \frac{\ln (t)}{1+t} d t$ Let $t=\frac{1}{y} \Rightarrow d t=\frac{-1}{y^{2}} d y$\ $f\left(\frac{1}{x}ight)=\int_{1}^{y} \frac{\ln 1 / y}{1+1 / y}\left(\frac{-1}{y^{2}}ight) d y=\int_{1}^{y} \frac{\ln y}{y(1+y)} d y$\ $f\left(\frac{1}{x}ight)=\int_{1}^{x} \frac{\ln (t)}{t(1+t)} d t$\ $f(x)+f\left(\frac{1}{x}ight)=\int_{1}^{x} \ln (t)\left[\frac{1}{t+1}+\frac{1}{t(t+1)}ight] d t$\ $=\int_{1}^{x} \frac{\ln (t)}{t} d t=\frac{1}{2}[\ln (x)]^{2}$\ $	herefore \quad f(e)+f\left(\frac{1}{e}ight)=\frac{1}{2}$.\