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arrow_back Given that $\int_{-\infty}^{\infty} \frac{e^{x}}{e^{2 x}-e^{-2 x}} d x$ is an improper integral, its convergence and its value can be found by considering

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(119,140 points)
in Mathematics
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Given that $\int_{-\infty}^{\infty} \frac{e^{x}}{e^{2 x}-e^{-2 x}} d x$ is an improper integral, its convergence and its value can be found by considering

A. $\lim _{t \rightarrow \infty} \int_{-t}^{t} \frac{e^{x}}{e^{2 x}-e^{-2 x}} d x$

B. $\lim _{t \rightarrow \infty} \int_{-t}^{2 t} \frac{e^{x}}{e^{2 x}-e^{-2 x}} d x$

C. $\lim _{t \rightarrow \infty} \int_{0}^{t}\left[\frac{e^{x}}{e^{2 x}-e^{-2 x}}-\frac{e^{-x}}{e^{2 x}-e^{-2 x}}\right] d x$

D. $\lim _{t \rightarrow \infty} \int_{0}^{t}\left[\frac{e^{x}}{e^{2 x}-e^{-2 x}}+\frac{e^{-x}}{e^{2 x}-e^{-2 x}}\right] d x$

E. $\lim _{t \rightarrow-\infty} \int_{t}^{1} \frac{e^{x}}{e^{2 x}-e^{-2 x}} d x$ and $\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{e^{x}}{e^{2 x}-e^{-2 x}} d x$

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