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

$$\int {\left[\frac{\mathbf{log}\mathbf{x}-1}{1+{\left(\mathbf{log}\mathbf{x}\right)}^2}\right]}^2\mathbf{dx}=\frac{\mathbf{x}}{{\left(\mathbf{log}\mathbf{x}\right)}^2+1}+\mathbf{C}$$

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

$$let\phi \left(x\right)=\frac{x}{1+{\left(lnx\right)}^2}wehave$$$${\phi }^{{}^{\prime }}\left(x\right)=\frac{1+{\left(lnx\right)}^2-x\left(2ln\left(x\right)\frac{1}{x}\right)}{{(1+{\left(ln\left(x\right)\right)}^2)}^2}$$$$=\frac{1+{\left(lnx\right)}^2-2ln\left(x\right)}{(1+{\left(lnx\right)}^2)}={\left(\frac{ln\left(x\right)-1}{1+{\left(lnx\right)}^2}\right)}^2\Rightarrow$$$$\int {\left(\frac{ln\left(x\right)-1}{1+{\left(lnx\right)}^2}\right)}^{2}dx=\frac{x}{1+{\left(lnx\right)}^2}+c.$$

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

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

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

$$\mathrm{Thanks}$$

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