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Question Number 95323 by ~blr237~ last updated on 24/May/20

$$\underset{x\to 0}{\lim }{\int }_x^{2x}\frac{ln(2+t)}{t}dt={\left(ln2\right)}^2$$

Answered by abdomathmax last updated on 24/May/20

$$\mathrm{\exists }\mathrm{c}\in ]\mathrm{x},2\mathrm{x}[/{\int }_{\mathrm{x}}^{2\mathrm{x}}\frac{\ln (2+\mathrm{t})}{\mathrm{t}}\mathrm{dt}=\ln (2+\mathrm{c}){\int }_{\mathrm{x}}^{2\mathrm{x}}\frac{\mathrm{dt}}{\mathrm{t}}$$$$=\ln (2+\mathrm{c})\ln \left(\frac{2\mathrm{x}}{\mathrm{x}}\right)(\mathrm{x}\to 0\Rightarrow \mathrm{c}\to 0\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}{\int }_{\mathrm{x}}^{2\mathrm{x}}\frac{\ln (2+\mathrm{t})}{\mathrm{t}}\mathrm{dt}=\ln \left(2\right).\ln \left(2\right)={\left(\ln 2\right)}^2$$

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