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Question Number 137203 by mathocean1 last updated on 31/Mar/21

$$\int \frac{ln(1+x)}{x}=?$$

Answered by Dwaipayan Shikari last updated on 31/Mar/21

$$\int \frac{log(1+x)}{x}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}\int \frac{x^{n-1}}{n}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{x}^n}{n^2}$$

Answered by Ar Brandon last updated on 31/Mar/21

$$\mathcal{I}={\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}={\int }_0^1\frac{1}{\mathrm{x}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{(\mathrm{n}+1)}\mathrm{dx}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}\mathrm{dx}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{{\mathrm{n}}^2}$$$$=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2\mathrm{n}+1)}^2}-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^2}=2\times \frac{3}{4}\zeta \left(2\right)-\zeta \left(2\right)=\frac{1}{2}\zeta \left(2\right)=\frac{{\pi }^2}{12}$$

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