Question Number 79531 by jagoll last updated on 26/Jan/20
$$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:? \\ $$
Commented by john santu last updated on 26/Jan/20
$${remember}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\pi\:{ln}\left(\mathrm{2}\right) \\ $$$${let}\:{u}=\frac{\mathrm{1}}{{x}}\Rightarrow{x}=\frac{\mathrm{1}}{{u}} \\ $$$${dx}=−\frac{{du}}{{u}^{\mathrm{2}} }\:\Rightarrow{x}=\mathrm{0},\:{u}=\infty\:{and}\: \\ $$$${x}=\mathrm{1}\:,\:{u}=\mathrm{1}\: \\ $$$$=\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{{ln}\left({u}+\frac{\mathrm{1}}{{u}}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}\left(\frac{−{du}}{{u}^{\mathrm{2}} }\right)={I} \\ $$$${I}=\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{{ln}\left({u}+\frac{\mathrm{1}}{{u}}\right){du}}{\mathrm{1}+{u}^{\mathrm{2}} }+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{ln}\left({u}+\frac{\mathrm{1}}{{u}}\right){du}}{\mathrm{1}+{u}^{\mathrm{2}} }− \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{ln}\left({u}+\frac{\mathrm{1}}{{u}}\right){du}}{\mathrm{1}+{u}^{\mathrm{2}} }\: \\ $$$$\mathrm{2}{I}=\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{ln}\left({u}+\frac{\mathrm{1}}{{u}}\right){du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\mathrm{2}{I}=\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{ln}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du}−\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{ln}\left({u}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$\mathrm{2}{I}=\:\pi{ln}\left(\mathrm{2}\right)−\mathrm{0}\:\Rightarrow{I}=\:\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}.\: \\ $$$$ \\ $$$$ \\ $$