Question-138206

Question Number 138206 by Bekzod Jumayev last updated on 11/Apr/21

Commented by Bekzod Jumayev last updated on 11/Apr/21

$${Help}? \\ $$

Answered by mr W last updated on 11/Apr/21

=∫x^(1/2) x^(1/(2×3)) x^(1/(2×3×4)) x^(1/(2×3×4×5)) ...dx =∫x^((1/2)+(1/(2×3))+(1/(2×3×4))+(1/(2×3×4×5))+...) dx =∫x^((1+1+(1/(2!))+(1/(3!))+(1/(4!))+(1/(5!))+...)−2) dx =∫x^(e−2) dx =(x^(e−1) /(e−1))+C

$$=\int{x}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}} {x}^{\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}} {x}^{\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}}} …{dx} \\ $$$$=\int{x}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}}+…} {dx} \\ $$$$=\int{x}^{\left(\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}+\frac{\mathrm{1}}{\mathrm{5}!}+…\right)−\mathrm{2}} {dx} \\ $$$$=\int{x}^{{e}−\mathrm{2}} {dx} \\ $$$$=\frac{{x}^{{e}−\mathrm{1}} }{{e}−\mathrm{1}}+{C} \\ $$

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Question-138206

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