For $x>0$, let $f(x) = \begin{align*} \int_1^x \frac{\log t}{1+t} dt \end{align*}$. Then $f(x) + f(\frac{1}{x})$ is equal to :

Question

( A ) $\frac{1}{4} (\log x)^2$
( B ) $\frac{1}{2} (\log x)^2$
( C ) $\log x$
( D ) $\frac{1}{4} \log x^2$