Differentiation Questions

Question Number 156864 by mnjuly1970 last updated on 16/Oct/21

$$\\$$$$\:\:\:\:\phi\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\:\left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)}{\mathrm{1}+\:{x}^{\:\mathrm{2}} }\:{dx}\:= \\$$$$\:\:{proof}\:: \\$$$$\:\:\:\:\phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\left(\mathrm{1}−{x}\:\right)}{\mathrm{1}+{x}^{\:\mathrm{2}} }{dx}\:+\:\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right) \\$$$$\:\:\:\:....\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\:\left(\:\mathrm{1}−{x}\:\right)}{\mathrm{1}+{x}^{\:\mathrm{2}} }{dx} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{{x}={tan}\left({t}\right)} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\:{cos}\left({t}\right)−{sin}\left({t}\right)\right){dt}−\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({t}\right)\right){dt} \\$$$$\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}\:\right){dt}\:+\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left(\frac{\pi}{\mathrm{4}}\:−{t}\right)\right){dt}−\frac{\mathrm{G}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right) \\$$$$\:\:\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:{ln}\left(\mathrm{2}\right)−\frac{\mathrm{G}}{\mathrm{2}}\:−\frac{\mathrm{G}}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\:{ln}\left(\mathrm{2}\right)=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)−\mathrm{G} \\$$$$\:\:\:\:\phi\:=\:\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:−\:\mathrm{G}\:\:\:\:\:\:\blacksquare\:{m}.{n} \\$$

Commented by mnjuly1970 last updated on 16/Oct/21

$$\:{God}\:{keep}\:{you}\:{sir} \\$$$${puissant}\:.{grateful} \\$$

Commented by puissant last updated on 16/Oct/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)+{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$=\:{Q}+{K} \\$$$${K}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:;\:{x}={tanu}\rightarrow{dx}=\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right){du} \\$$$$\Rightarrow\:{K}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tanu}\right){du}\:;\:{u}=\frac{\pi}{\mathrm{4}}−{t}\rightarrow{du}=−{dt} \\$$$$\Rightarrow\:{K}=\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} {ln}\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{t}\right)\right)\left(−{dt}\right) \\$$$$\Rightarrow\:{K}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}−{tant}}{\mathrm{1}+{tant}}\right){dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{\mathrm{2}}{\mathrm{1}+{tant}}\right){dt} \\$$$$\Rightarrow\:\mathrm{2}{K}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\mathrm{2}\:{dt}\:\Rightarrow\:{K}=\:\frac{\pi}{\mathrm{8}}{ln}\mathrm{2} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\because\:\:\:{K}\:=\:\frac{\pi}{\mathrm{8}}{ln}\mathrm{2}.. \\$$

Commented by puissant last updated on 16/Oct/21

$${Great}\:,\:{Good}\:{sir}\:{Mnjuly}\mathrm{1970}. \\$$

Commented by mnjuly1970 last updated on 16/Oct/21

$${thanks}\:{alot}\:... \\$$