Question

The integral \( \int_{0}^{1} \frac{1}{{ }_{7}\left[\frac{1}{x}\right]} d x \), what n:

Answer 1

Correct option is (A) \( 1+6 \log _{ e }( 6 / 7) \) \[ \begin{aligned} & \int_{0}^{1} \frac{1}{\left[\frac{1}{7^{2}}\right]} dx =-\int_{1}^{0} \frac{1}{7^{\left.\frac{1}{x}\right]}} dx \\ = & (-1)\left[\int_{1}^{1 / 2} \frac{1}{7} dx +\int_{1 / 2}^{1 / 3} \frac{1}{7^{2}} dx +\int_{1 / 3}^{1 / 4} \frac{1}{7^{3}} dx +\ldots \ldots \infty\right] \\ = & \left(\frac{1}{7}+\frac{1}{2.7^{2}}+\frac{1}{3.7^{3}}+\ldots \infty\right)-\left(\frac{1}{7.2}+\frac{1}{7^{2} \cdot 3}+\frac{1}{7^{2} \cdot 4} \ldots \infty\right) \\ = & -\ln \left(1-\frac{1}{7}\right)-7\left(\frac{1}{7^{2} .2}+\frac{1}{7^{3} \cdot 3}+\frac{1}{7^{4} \cdot 4}+\ldots . . \infty\right) \\ & {\left[\text { as } \ln (1+x)= x -\frac{ x ^{2}}{2}+\frac{ x ^{3}}{3}-\frac{ x ^{4}}{4} \ldots . \infty\right] } \\ = & -\ln \frac{6}{7}-7\left(-\ln \left(1-\frac{1}{7}\right)-\frac{1}{7}\right) \\ = & 6 \ln \frac{6}{7}+1 \end{aligned} \]