$$\int_{-1}^{1} \frac{e^x}{e^x-1} \,dx = \lim_{z \to 0} (\int_{-1}^{z} \frac{e^x}{e^x-1} \,dx + \int_{z}^{1} \frac{e^x}{e^x-1} \,dx )$$
$$=\lim_{z \to 0} (\int_{-1}^{z} \frac{1}{u}\,du + \int_{z}^{1} \frac{1}{u}\,du) $$
$$=\lim_{z \to 0} (\ln|u||^{z}_{-1} + \ln|u||^{1}_{z})$$
$$=\lim_{z \to 0} (\ln|{e^x-1}||^{z}_{-1} + \ln|{e^x-1}||^{1}_{z})$$
$$=\lim_{z \to 0} (\ln{(e^z-1)} - \ln{(1-e^{-1})} + \ln{(e-1)} - \ln{(e^{z}-1)})$$
$$=\ln \frac{e-1}{1-e^{-1}}=1$$
ผิดตรงไหนอะครับ?
