# ISI2015-MMA-80

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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
recategorized
0
$\ln |x + c|$ ?
1
B ? by applying the telescoping technique.
0
B is the answer

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