Find-1-1-x-9-dx-2-1-x-9-1-dx-

Question Number 221089 by shunmisaki007 last updated on 24/May/25

$$\mathrm{Find}\: \\ $$$$\left(\mathrm{1}\right)\:\int\frac{\mathrm{1}}{{x}^{\mathrm{9}} }{dx} \\ $$$$\left(\mathrm{2}\right)\:\int\frac{\mathrm{1}}{{x}^{\mathrm{9}} +\mathrm{1}}{dx}. \\ $$

Answered by MrGaster last updated on 24/May/25

For (1): ∫(1/x^9 )dx=∫x^(−9) dx =(x^(−8) /(−8))+C =−(1/(8x^8 ))+C

$$\mathrm{For}\:\left(\mathrm{1}\right):\:\int\frac{\mathrm{1}}{{x}^{\mathrm{9}} }{dx}=\int{x}^{−\mathrm{9}} {dx} \\ $$$$=\frac{{x}^{−\mathrm{8}} }{−\mathrm{8}}+{C} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{8}} }+{C} \\ $$

Commented by MrGaster last updated on 24/May/25

Answered by SdC355 last updated on 24/May/25

∫ dz Σ_(h=0) ^∞ (−)^h z^(9h) =Σ_(h=0) ^∞ (((−1)^h )/(9h+1))z^(9h+1) , −1<z≤1

$$\int\:\:\mathrm{d}{z}\:\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\right)^{{h}} {z}^{\mathrm{9}{h}} =\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{h}} }{\mathrm{9}{h}+\mathrm{1}}{z}^{\mathrm{9}{h}+\mathrm{1}} ,\:−\mathrm{1}<{z}\leq\mathrm{1} \\ $$

Commented by MathematicalUser2357 last updated on 30/May/25

실수를 만들었네요. (-)가 아니라 (-1)입니다.

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