JEE Main 2022 — Mathematics Calculus

$$I={\int }_{0}^{1}\frac{dx}{{7}^{[\frac{1}{x}]}}={\int }_{0}^{1}{(\frac{1}{7})}^{[\frac{1}{x}]}dx$$I=∫01​7[x1​]dx​=∫01​(71​)[x1​]dx

Let $\displaystyle \frac{1}{7}=k\Rightarrow I={\int }_{0}^{1}{k}^{[\frac{1}{x}]}dx$71​=k⇒I=∫01​k[x1​]dx

$$I={\int }_{0}^{1}{k}^{[\frac{1}{x}]}dx$$I=∫01​k[x1​]dx

$$\Rightarrow I={\int }_{\frac{1}{2}}^{1}{k}^{[\frac{1}{x}]}dx+{\int }_{\frac{1}{3}}^{\frac{1}{2}}{k}^{[\frac{1}{x}]}dx+{\int }_{\frac{1}{4}}^{\frac{1}{3}}{k}^{[\frac{1}{x}]}dx+\cdots$$⇒I=∫21​1​k[x1​]dx+∫31​21​​k[x1​]dx+∫41​31​​k[x1​]dx+⋯

$$\Rightarrow I=k(1-\frac{1}{2})+{k}^{2}(\frac{1}{2}-\frac{1}{3})+{k}^{3}(\frac{1}{3}-\frac{1}{4})+\cdots$$⇒I=k(1−21​)+k2(21​−31​)+k3(31​−41​)+⋯

$$\Rightarrow I=(k+\frac{{k}^{2}}{2}+\frac{{k}^{3}}{3}+\cdots )-\frac{1}{k}(\frac{{k}^{2}}{2}+\frac{{k}^{3}}{3}+\cdots )$$⇒I=(k+2k2​+3k3​+⋯)−k1​(2k2​+3k3​+⋯)

$$\Rightarrow I=-\mathrm{ln}(1-k)-\frac{1}{k}[-\mathrm{ln}(1-k)-k]$$⇒I=−ln(1−k)−k1​[−ln(1−k)−k]

$$\Rightarrow I=-\mathrm{ln}\frac{6}{7}-7[-\mathrm{ln}\frac{6}{7}-\frac{1}{7}](\because k=\frac{1}{7})$$⇒I=−ln76​−7[−ln76​−71​](∵k=71​)

$$\therefore I=1+6\mathrm{ln}\frac{6}{7}$$∴I=1+6ln76​