$$\lim_{B\to\infty} \int_{1}^{B} \frac{e^x}{e^{2x}-1} dx$$
This equals
$$=\lim_{B\to\infty} \frac{1}{2}\big(\ln |e^{x} - 1| - \ln |e^{x} + 1| \big) \bigg|_{1}^{B}$$
$$= \lim_{b\to\infty} \frac{1}{2}\bigg(\ln \bigg|\frac{e^x -1}{e^x + 1}\bigg| \bigg) \bigg|_{1}^{B}$$
$$= \frac{1}{2}\bigg(\ln 1 - \ln\bigg|\frac{e -1}{e + 1}\bigg| \bigg)$$
$$= -\frac{1}{2}\ln\bigg|\frac{e -1}{e + 1}\bigg|$$
Answer is undefined.