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Question Number 130907 by greg_ed last updated on 30/Jan/21

∫_0 ^( 1)  ((ln (1+x+x^2 ))/x) dx

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)}{{x}}\:{dx} \\ $$

Commented by benjo_mathlover last updated on 30/Jan/21

այս հարցն արդեն պատասխանվել է

Commented by benjo_mathlover last updated on 30/Jan/21

qn 130676

$$\mathrm{qn}\:\mathrm{130676} \\ $$

Answered by Dwaipayan Shikari last updated on 30/Jan/21

∫_0 ^1 ((log(1−x^3 ))/x)−((log(1−x))/x)dx  =−Σ_(n=1) ^∞ ∫_0 ^1 (x^(3n−1) /n)+Σ_(n=1) ^∞ ∫_0 ^1 (x^(n−1) /n)dx  =−(1/3)Σ_(n=1) ^∞ (1/n^2 )+Σ_(n=1) ^∞ (1/n^2 )=ζ(2)(1−(1/3))=(π^2 /9)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{x}^{\mathrm{3}} \right)}{{x}}−\frac{{log}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}{n}−\mathrm{1}} }{{n}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}−\mathrm{1}} }{{n}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\zeta\left(\mathrm{2}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{9}} \\ $$

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