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Question Number 31517 by abdo imad last updated on 09/Mar/18
find ∫_(−1) ^1 (dx/( (√(1+x)) +(√(1−x)))) .
find∫−11dx1+x+1−x.
Commented by abdo imad last updated on 12/Mar/18
let put I(ξ) =∫_(−1+ξ) ^(1+ξ) (dx/( (√(1+x)) +(√(1−x)) )) we have I=lim_(ξ→0) I(ξ) but I(ξ)= ∫_(−1+ξ) ^(1+ξ) (((√(1+x)) −(√(1−x)) )/(2x))dx =(1/2)( ∫_(−1+ξ) ^(1+ξ) ((√(1+x))/x)dx −∫_(−1+ξ) ^(1+ξ) ((√(1−x))/x) dx ) ch.(√(1+x)) =t ⇒1+x=t^2 ⇒x=t^2 −1 give ∫_(−1+ξ) ^(1+ξ) ((√(1+x))/x)dx= ∫_(√ξ) ^(√(2+ξ)) (t/(t^2 −1)) (2t)dt = 2∫_(√ξ) ^(√(2+ξ)) ((t^2 −1 +1)/(t^2 −1))dt=2((√(2+ξ)) −(√ξ)) + ∫_(√ξ) ^(√(2+ξ)) ((1/(t−1)) −(1/(t+1)))dt =2((√(2+ξ)) −(√ξ)) + [ln∣((t−1)/(t+1))∣]_(√ξ) ^(√(2+ξ)) =2((√(2+ξ)) −(√ξ) ) +ln∣(((√(2+ξ)) −1)/( (√(2+ξ)) +1))∣ −ln∣(((√ξ) −1)/( (√ξ) +1))∣ →_(ξ→0) 2(√2) +ln((((√2) −1)/( (√2) +1))) and ch.(√(1−x)) =t⇒1−x=t^2 ⇒x=1−t^2 give ∫_(−1+ξ) ^(1+ξ) ((√(1−x))/x)dx = ∫_(√(2−ξ)) ^(√(−ξ)) (t/(1−t^2 )) (−2t)dt = 2 ∫_(√(2−ξ)) ^(√(−ξ)) ((t^2 −1+1)/(t^2 −1))dt= 2((√(−ξ)) −(√(2−ξ))) +∫_(√(2−ξ)) ^(√(−ξ)) ((1/(t−1)) −(1/(t+1)))dt =2((√(−ξ)) −(√(2−ξ))) +[ln∣((t−1)/(t+1))∣]_(√(2−ξ)) ^(√(−ξ)) =2((√(−ξ)) −(√(2−ξ)) ) + ln ∣(((√(−ξ)) −1)/( (√(−ξ)) +1))∣ −ln∣(((√(2−ξ)) −1)/( (√(2−ξ)) +1))∣ →−2(√2) −ln((((√2) −1)/( (√2) +1))) ⇒ lim_(ξ→0) I(ξ) =(1/2)(2(√2) +ln((((√2) −1)/( (√2) +1))) +2(√(2 )) +ln((((√2) −1)/( (√2) +1)))) =2(√2) +ln((((√2) −1)/( (√2) +1))) .
letputI(ξ)=∫−1+ξ1+ξdx1+x+1−xwehaveI=limξ→0I(ξ)butI(ξ)=∫−1+ξ1+ξ1+x−1−x2xdx=12(∫−1+ξ1+ξ1+xxdx−∫−1+ξ1+ξ1−xxdx)ch.1+x=t⇒1+x=t2⇒x=t2−1give∫−1+ξ1+ξ1+xxdx=∫ξ2+ξtt2−1(2t)dt=2∫ξ2+ξt2−1+1t2−1dt=2(2+ξ−ξ)+∫ξ2+ξ(1t−1−1t+1)dt=2(2+ξ−ξ)+[ln∣t−1t+1∣]ξ2+ξ=2(2+ξ−ξ)+ln∣2+ξ−12+ξ+1∣−ln∣ξ−1ξ+1∣→ξ→022+ln(2−12+1)andch.1−x=t⇒1−x=t2⇒x=1−t2give∫−1+ξ1+ξ1−xxdx=∫2−ξ−ξt1−t2(−2t)dt=2∫2−ξ−ξt2−1+1t2−1dt=2(−ξ−2−ξ)+∫2−ξ−ξ(1t−1−1t+1)dt=2(−ξ−2−ξ)+[ln∣t−1t+1∣]2−ξ−ξ=2(−ξ−2−ξ)+ln∣−ξ−1−ξ+1∣−ln∣2−ξ−12−ξ+1∣→−22−ln(2−12+1)⇒limξ→0I(ξ)=12(22+ln(2−12+1)+22+ln(2−12+1))=22+ln(2−12+1).

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