For $x >0$ , let $f(x)= \int \limits_1^x \large\frac{log t}{1+t}$$dt$. Then $f(x)+f\bigg(\large\frac{1}{x}\bigg)$ is equal to:

Question

$\begin{array}{1 1} \frac{1}{4} (\log x)^2 \\ \frac{1}{2}(\log x)^2 \\ log\; x \\ \frac{1}{4} (\log x)^2 \end{array} $