The integral $\int_{0}^{1} \frac{1}{7^{[\frac{1}{x}]}} d x$ , where [.] denotes the greatest integer function is equal to

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1- $6 \log_{e} (\frac{6}{7})$

1+ $6 \log_{e} (\frac{6}{7})$

$\log_{e} (\frac{7}{6})$

1- $7 \log_{e} (\frac{6}{7})$

answer is A.

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Detailed Solution

$\begin{array}{l} \int_{0}^{1} \frac{1}{7^{[\frac{1}{x}]}} dx = - \int_{1}^{0} \frac{1}{7^{[\frac{1}{x}]}} dx \\ = (- 1) [\int_{1}^{1 / 2} \frac{1}{7} dx + \int_{1 / 2}^{1 / 3} \frac{1}{7^{2}} dx + \int_{1 / 3}^{1 / 4} \frac{1}{7^{3}} d x + \dots \dots \infty] \\ = (\frac{1}{7} + \frac{1}{2 \cdot 7^{2}} + \frac{1}{3 \cdot 7^{3}} + \dots \infty) - (\frac{1}{7 \cdot 2} + \frac{1}{7^{2} \cdot 3} + \frac{1}{7^{2} \cdot 4} \dots \infty) \\ = - \ln (1 - \frac{1}{7}) - 7 (\frac{1}{7^{2} \cdot 2} + \frac{1}{7^{3} \cdot 3} + \frac{1}{7^{4} \cdot 4} + \dots \dots \infty) \\ [as \ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} \dots \infty] \\ = - \ln \frac{6}{7} - 7 (- \ln (1 - \frac{1}{7}) - \frac{1}{7}) \\ = 6 \ln \frac{6}{7} + 1 \end{array}$