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Question Number 44604 by arvinddayama01@gmail.com last updated on 02/Oct/18

∫[((log x − 1)/(1+(log x)^2 ))]^2 dx = (x/((log x)^2 +1))+C

$$\int\left[\frac{\boldsymbol{\mathrm{log}}\:\boldsymbol{\mathrm{x}}\:\:−\:\:\mathrm{1}}{\mathrm{1}+\left(\boldsymbol{\mathrm{log}}\:\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} \boldsymbol{\mathrm{dx}}\:\:=\:\:\frac{\boldsymbol{\mathrm{x}}}{\left(\boldsymbol{\mathrm{log}}\:\boldsymbol{\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{1}}+\boldsymbol{\mathrm{C}} \\ $$

Commented by prof Abdo imad last updated on 02/Oct/18

let ϕ(x)=(x/(1+(lnx)^2 )) we have ϕ^′ (x)=((1+(lnx)^2 −x(2ln(x)(1/x)))/((1+(ln(x))^2 )^2 )) =((1+(lnx)^2 −2ln(x))/((1+(lnx)^2 ))) =(((ln(x)−1)/(1+(lnx)^2 )))^2 ⇒ ∫ (((ln(x)−1)/(1+(lnx)^2 )))^2 dx=(x/(1+(lnx)^2 )) +c .

$${let}\:\varphi\left({x}\right)=\frac{{x}}{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} }\:\:{we}\:{have} \\ $$$$\varphi^{'} \left({x}\right)=\frac{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} −{x}\left(\mathrm{2}{ln}\left({x}\right)\frac{\mathrm{1}}{{x}}\right)}{\left(\mathrm{1}+\left({ln}\left({x}\right)\right)^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} −\mathrm{2}{ln}\left({x}\right)}{\left(\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} \right)}\:=\left(\frac{{ln}\left({x}\right)−\mathrm{1}}{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\int\:\:\left(\frac{{ln}\left({x}\right)−\mathrm{1}}{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} {dx}=\frac{{x}}{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} }\:+{c}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

(d/dx)((u/v))=((v(du/dx)−u(dv/dx))/v^2 ) ∫(((^ lnx)^2 +1−2lnx)/({1+(lnx)^2 }^2 ))dx ∫(({1+(lnx)^2 }(dx/dx)−x.(d/dx){1+(lnx)^2 } )/({1+(lnx)^2 ))dx ∫(d/dx){(x/(1+(lnx)^2 ))}dx ∫d{(x/(1+(lnx)^2 ))} (x/(1+(lnx)^2 ))+c

$$\frac{{d}}{{dx}}\left(\frac{{u}}{{v}}\right)=\frac{{v}\frac{{du}}{{dx}}−{u}\frac{{dv}}{{dx}}}{{v}^{\mathrm{2}} } \\ $$$$\int\frac{\left(^{} {lnx}\right)^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{lnx}}{\left\{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} \right\}^{\mathrm{2}} }{dx} \\ $$$$\int\frac{\left\{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} \right\}\frac{{dx}}{{dx}}−{x}.\frac{{d}}{{dx}}\left\{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} \right\}\:}{\left\{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} \right.}{dx} \\ $$$$\int\frac{{d}}{{dx}}\left\{\frac{{x}}{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} }\right\}{dx} \\ $$$$\int{d}\left\{\frac{{x}}{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} }\right\} \\ $$$$\frac{{x}}{\mathrm{1}+\left({lnx}\right)^{\mathrm{2}} }+{c} \\ $$

Commented by arvinddayama01@gmail.com last updated on 02/Oct/18

Thanks

$$\mathrm{Thanks} \\ $$

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