Question Number 166373 by mnjuly1970 last updated on 19/Feb/22
![prove 𝛗=∫_0 ^( 1) (( ln^( 2) (1−x^( 2) ) )/x^( 2) ) dx =(π^( 2) /3) −4ln^( 2) (2) −−− solution (technical method) −−− 𝛗= ∫_0 ^( 1) ln^( 2) (1−x^( 2) )d(1−(1/x)) = [(1−(1/x))ln^( 2) (1−x^( 2) )]_0 ^1 +4∫_0 ^( 1) (1−(1/x))((xln(1−x^( 2) ))/(1−x^( 2) ))dx = −4∫_0 ^( 1) ((ln(1−x^( 2) ))/(1+x)) dx = −4∫_0 ^( 1) ((ln(1+x))/(1+x))dx −4∫_0 ^( 1) ((ln(1−x)dx)/(1+x)) = −2ln^( 2) (2) −4 ( −(π^( 2) /(12)) +(1/2)ln^( 2) (2)) ∴ 𝛗= (π^( 2) /3) −4ln^( 2) (2) ■ m.n](https://www.tinkutara.com/question/Q166373.png)
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove}\:\: \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)\:}{{x}^{\:\mathrm{2}} }\:{dx}\:=\frac{\pi^{\:\mathrm{2}} }{\mathrm{3}}\:−\mathrm{4}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:−−−\:\:{solution}\:\left({technical}\:{method}\right)\:−−− \\ $$$$\:\:\:\:\boldsymbol{\phi}=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{\:\mathrm{2}} \right){d}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\:\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\frac{{xln}\left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)}{\mathrm{1}−{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:=\:−\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)}{\mathrm{1}+{x}}\:{dx}\: \\ $$$$\:\:\:\:\:\:\:=\:−\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx}\:−\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right){dx}}{\mathrm{1}+{x}} \\ $$$$\:\:\:\:\:\:=\:−\mathrm{2}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right)\:−\mathrm{4}\:\left(\:−\frac{\pi^{\:\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{3}}\:−\mathrm{4}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\blacksquare\:\:{m}.{n}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$