If $f(x)=\int\limits_1^x \frac{\log t}{1+t} d t$, then $f(x)+f\left(\frac{1}{x}\right)$ is equal to

$\frac{1}{2}\left(\log _e x\right)^2$

We have,

$f(x)=\int\limits_1^x \frac{\log t}{1+t} d t$

∴  $f\left(\frac{1}{x}\right)=\int\limits_1^{1 / x} \frac{\log t}{1+t} d t$

$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_1^x \frac{\log \left(\frac{1}{u}\right)}{1+\frac{1}{u}} \times\left(-\frac{1}{u^2}\right)$, where $t=\frac{1}{u}$

$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_1^x \frac{\log u}{u(1+u)} d u$

$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_1^x \frac{\log t}{t(1+t)} d t$

∴  $f(x)+f\left(\frac{1}{x}\right)$

$=\int\limits_1^x \frac{\log t}{1+t} d t+\int\limits_1^x \frac{\log t}{t(1+t)} d t$

$=\int\limits_1^x \frac{\log t}{t} d t=\int\limits_1^x \log t d(\log t)=\left[\frac{(\log t)^2}{2}\right]_1^x=\frac{\left(\log _e x\right)^2}{2}$