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Let $0 < \alpha < \beta < 1$. Then $$ \Sigma_{k=1}^{\infty} \int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{1}{1+x} dx$$ is equal to

  1. $\log_e \frac{\beta}{\alpha}$
  2. $\log_e \frac{1+ \beta}{1 + \alpha}$
  3. $\log_e \frac{1+\alpha }{1+ \beta}$
  4. $\infty$
in Calculus
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$\ln |x + c| $ ?
1
B ? by applying the telescoping technique.
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B is the answer

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