Let us apply this method to the integral $$\begin{equation} I(x)={\cal P}\int_{-1}^{+1}dt\frac{e^t}{t}. \tag{23} \end{equation}$$ The integrand diverges at $x=t=0$. We rewrite it using Eq. (21) as $$\begin{equation} {\cal P}\int_{-1}^{+1}dt\frac{e^t}{t}=\int_{-1}^{+1}\frac{e^t-1}{t}, \tag{24} \end{equation}$$ since $e^x=e^0=1$. With Eq. (22) we have then $$\begin{equation} \int_{-1}^{+1}\frac{e^t-1}{t}\approx \sum_{i=1}^{N}\omega_i\frac{e^{t_i}-1}{t_i}. \tag{25} \end{equation}$$