Question Number 167882 by LEKOUMA last updated on 28/Mar/22

Answered by ArielVyny last updated on 29/Mar/22
![t=(√((1−x)/(1+x)))→dt=((−(2/((1+x)^2 )))/(2t))dx=−(1/((1+x)^2 ))(1/t)dx t^2 =((1−x)/(1+x))→1−t^2 =x(t^2 +1)→x=((1−t^2 )/(1+t^2 )) dt=−(1/((((1+t^2 )/(1+t^2 ))+((1−t^2 )/(1+t^2 )))t))dx=−.((1+t^2 )/t)dx I=−∫((1+t^2 )/(1−t^2 )).t.(t/(1+t^2 ))dt=∫((−1−t^2 +1)/(1−t^2 ))dt I=t−∫(1/(1−t^2 ))dt t=sinu→dt=cosudu I=t−∫(1/(cos^2 u))cosudu=(√((1−x)/(1+x)))+∫(1/(cosu))du I=(√((1−x)/(1+x)))+∫((cosu)/((1−sin^2 u)))du =.....−(1/2)(−∫−((cosu)/(1−sinu))+∫((cosu)/(1+sinu))du) =.....−(1/2)[−ln(1−sinu)+ln(1+sinu)] =.....−(1/2)ln(((1+sinu)/(1−sinu))) =.....−(1/2)ln(((1+t)/(1−t)))=(√(((1−x)/(1+x))−))(1/2)ln(((1+(√((1−x)/(1+x))))/(1−(√((1−x)/(1+x)))))) I=(√(((1−x)/(1+x))−))(1/2)ln(((((√(1+x))+(√(1−x)))/( (√(1+x))))/(((√(1+x))−(√(1−x)))/( (√(1+x)))))) =(√((1−x)/(1+x)))−(1/2)ln((((√(1+x))+(√(1−x)))/( (√(1+x))−(√(1−x))))) =(√((1−x)/(1+x)))−(1/2)ln[((((√(1+x))+(√(1−x)))^2 )/(((√(1+x)))^2 −((√(1−x)))^2 ))] =(√((1−x)/(1+x)))−(1/2)ln[((1+(√(1−x^2 )))/x)]](https://www.tinkutara.com/question/Q167886.png)
Commented by peter frank last updated on 28/Mar/22

Commented by MJS_new last updated on 29/Mar/22
![this is wrong. what you do in the 1^(st) and 3^(rd) lines makes no sense. t=((√(1−x))/( (√(1+x)))) ⇔ x=((1−t^2 )/(1+t^2 )) dx=(dt/((d/dx)[((√(1−x))/( (√(1+x))))]))=(dt/(−(1/( (√(1−x))(√((1+x)^3 ))))))=−(√(1−x))(√((1+x)^3 ))dt ⇒ ∫((√(1−x))/(x(√(1+x))))dx=∫((√(1−x))/(x(√(1+x))))×(−(√(1−x))(√((1+x)^3 ))dt)= =∫(((x−1)(x+1))/x)dt=∫((((−2t^2 )/(t^2 +1))×(2/(t^2 +1)))/(−((t^2 −1)/(t^2 +1))))dt= =4∫(t^2 /((t^2 −1)(t^2 +1)))dt](https://www.tinkutara.com/question/Q167913.png)
Commented by ArielVyny last updated on 29/Mar/22

Answered by MJS_new last updated on 28/Mar/22
![∫((√(1−x))/(x(√(1+x))))dx= [t=((√(1−x))/( (√(1+x)))) → dx=−(√(1−x))(√((1+x)^3 ))dt] =4∫(t^2 /(t^4 −1))dt=∫((2/(t^2 +1))+(1/(t−1))−(1/(t+1)))dt= =2arctan t +ln ((t−1)/(t+1)) = =2arctan ((√(1−x))/( (√(1+x)))) +ln ∣((1−(√(1−x^2 )))/x)∣ +C](https://www.tinkutara.com/question/Q167895.png)