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Question Number 166373 by mnjuly1970 last updated on 19/Feb/22
proveϕ=∫01ln2(1−x2)x2dx=π23−4ln2(2)−−−solution(technicalmethod)−−−ϕ=∫01ln2(1−x2)d(1−1x)=[(1−1x)ln2(1−x2)]01+4∫01(1−1x)xln(1−x2)1−x2dx=−4∫01ln(1−x2)1+xdx=−4∫01ln(1+x)1+xdx−4∫01ln(1−x)dx1+x=−2ln2(2)−4(−π212+12ln2(2))∴ϕ=π23−4ln2(2)◼m.n
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![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](Q166373.png)