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Question Number 146035 by KONE last updated on 10/Jul/21

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

Answered by KONE last updated on 10/Jul/21

$$please$$

Answered by puissant last updated on 10/Jul/21

$$\frac{1}{1+\mathrm{x}}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$$$$\Rightarrow \ln (1+\mathrm{x})=\int \underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\mathrm{dx}$$$$\Rightarrow \ln (1+\mathrm{x})=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$$$$\mathrm{x}=0\Rightarrow \ln (1+\mathrm{x})=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$$$$\Rightarrow \frac{\ln (1+\mathrm{x})}{\mathrm{x}}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}+1}$$$$\Rightarrow \int \frac{\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{1}{\mathrm{n}+1}\int {\mathrm{x}}^{\mathrm{n}}\mathrm{dx}$$$$\Rightarrow \mathrm{I}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}+\mathrm{k}..$$

Commented by KONE last updated on 12/Jul/21

$$thanks...$$$$svpestpossibledefairesansutiliserleDL?$$

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