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Question Number 68879 by mathmax by abdo last updated on 16/Sep/19

$$letI={\int }_0^1\frac{x}{ln(1+x)}dxdetermineaapproximatevalueofI$$

Commented by Abdo msup. last updated on 21/Sep/19

$$thereaerrorofcalculusintheansweriwillhid$$$$thispost...$$

Commented by mathmax by abdo last updated on 22/Sep/19

$$wehave{ln}^{{}^{\prime }}(1+x)=\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n\Rightarrow$$$$ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{n+1}}{n+1}+c(c=0)\Rightarrow$$$$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...\Rightarrow x-\frac{x^2}{2}⩽ln(1+x)⩽x\Rightarrow$$$$\frac{1}{x}⩽\frac{1}{ln(1+x)}⩽\frac{1}{x-\frac{x^2}{2}}\Rightarrow 1⩽\frac{x}{ln(1+x)}⩽\frac{1}{1-\frac{x}{2}}\Rightarrow$$$$1⩽{\int }_0^1\frac{x}{ln(1+x)}dx⩽{\int }_0^1\frac{dx}{1-\frac{x}{2}}wehave$$$${\int }_0^1\frac{dx}{1-\frac{x}{2}}={\int }_0^1\frac{2dx}{2-x}=-2{\int }_0^1\frac{dx}{x-2}=-2{[ln\mid x-2\mid ]}_0^1=-2(-ln2)$$$$=2ln\left(2\right)\Rightarrow 1⩽I⩽2ln\left(2\right)so{v}_0=\frac{1+2ln\left(2\right)}{2}isaapproximate$$$$valueofI.$$

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