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Question Number 200569 by Rupesh123 last updated on 20/Nov/23

Answered by MM42 last updated on 20/Nov/23

$$★e=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}\Rightarrow \underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=e-2$$$$\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{\sqrt{x...}}}}}=x^{\frac{1}{2}}\times x^{\frac{1}{6}}\times x^{\frac{1}{24}}\times ...$$$$=x^{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...}=x^{e-2}$$$$\Rightarrow I=\int x^{e-2}dx=\frac{x^{e-1}}{e-1}+c✓$$$$$$

Commented by Rupesh123 last updated on 20/Nov/23

Nice one, sir!

Commented by MM42 last updated on 20/Nov/23

$$\underset{―}{\underbrace{⋚}}$$

Commented by Harnada last updated on 21/Nov/23

$$Q.1$$

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