Considering four sub-intervals, the value of $\int_{0}^{1} \frac{1}{1+x} d x$ by Trapezoidal rule, is

\begin{array}{c|l|l} \hline \boldsymbol{i} & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{y}_{\boldsymbol{i}}=\frac{\mathbf{1}}{\mathbf{1 +} \boldsymbol{x}_{\boldsymbol{i}}} \\ \hline 0 & 0 & 1 \\ 1 & 0.25 & 0.8 \\ 2 & 0.5 & 0.67 \\ 3 & 0.75 & 0.571 \\ 4 & 1 & 0.5 \\ \hline \end{array} Trapezoidal rule gives $\begin{aligned} \int_{0}^{1} \frac{1}{1+x} d x=& \frac{h}{2}\left[y_{0}+2\left(y_{1}+y_{2}+y_{3}\right)+y_{4}\right] \\=& \frac{1-0}{2 \times 4}[1+2(0.8+0.67\\ & \\=& 0.6977 \end{aligned}$