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Calculate-I-1-x-1-x-1-x-dx-Indication-poser-t-1-x-1-x-




Question Number 167882 by LEKOUMA last updated on 28/Mar/22
Calculate  I=โˆซ(1/x)((โˆš((1โˆ’x)/(1+x))))dx  Indication poser t=(โˆš((1โˆ’x)/(1+x)))
CalculateI=โˆซ1x(1โˆ’x1+x)dxIndicationposert=1โˆ’x1+x
Answered by ArielVyny last updated on 29/Mar/22
t=(โˆš((1โˆ’x)/(1+x)))โ†’dt=((โˆ’(2/((1+x)^2 )))/(2t))dx=โˆ’(1/((1+x)^2 ))(1/t)dx  t^2 =((1โˆ’x)/(1+x))โ†’1โˆ’t^2 =x(t^2 +1)โ†’x=((1โˆ’t^2 )/(1+t^2 ))  dt=โˆ’(1/((((1+t^2 )/(1+t^2 ))+((1โˆ’t^2 )/(1+t^2 )))t))dx=โˆ’.((1+t^2 )/t)dx  I=โˆ’โˆซ((1+t^2 )/(1โˆ’t^2 )).t.(t/(1+t^2 ))dt=โˆซ((โˆ’1โˆ’t^2 +1)/(1โˆ’t^2 ))dt  I=tโˆ’โˆซ(1/(1โˆ’t^2 ))dt  t=sinuโ†’dt=cosudu  I=tโˆ’โˆซ(1/(cos^2 u))cosudu=(โˆš((1โˆ’x)/(1+x)))+โˆซ(1/(cosu))du  I=(โˆš((1โˆ’x)/(1+x)))+โˆซ((cosu)/((1โˆ’sin^2 u)))du    =.....โˆ’(1/2)(โˆ’โˆซโˆ’((cosu)/(1โˆ’sinu))+โˆซ((cosu)/(1+sinu))du)    =.....โˆ’(1/2)[โˆ’ln(1โˆ’sinu)+ln(1+sinu)]    =.....โˆ’(1/2)ln(((1+sinu)/(1โˆ’sinu)))    =.....โˆ’(1/2)ln(((1+t)/(1โˆ’t)))=(โˆš(((1โˆ’x)/(1+x))โˆ’))(1/2)ln(((1+(โˆš((1โˆ’x)/(1+x))))/(1โˆ’(โˆš((1โˆ’x)/(1+x))))))  I=(โˆš(((1โˆ’x)/(1+x))โˆ’))(1/2)ln(((((โˆš(1+x))+(โˆš(1โˆ’x)))/( (โˆš(1+x))))/(((โˆš(1+x))โˆ’(โˆš(1โˆ’x)))/( (โˆš(1+x))))))    =(โˆš((1โˆ’x)/(1+x)))โˆ’(1/2)ln((((โˆš(1+x))+(โˆš(1โˆ’x)))/( (โˆš(1+x))โˆ’(โˆš(1โˆ’x)))))     =(โˆš((1โˆ’x)/(1+x)))โˆ’(1/2)ln[((((โˆš(1+x))+(โˆš(1โˆ’x)))^2 )/(((โˆš(1+x)))^2 โˆ’((โˆš(1โˆ’x)))^2 ))]     =(โˆš((1โˆ’x)/(1+x)))โˆ’(1/2)ln[((1+(โˆš(1โˆ’x^2 )))/x)]
t=1โˆ’x1+xโ†’dt=โˆ’2(1+x)22tdx=โˆ’1(1+x)21tdxt2=1โˆ’x1+xโ†’1โˆ’t2=x(t2+1)โ†’x=1โˆ’t21+t2dt=โˆ’1(1+t21+t2+1โˆ’t21+t2)tdx=โˆ’.1+t2tdxI=โˆ’โˆซ1+t21โˆ’t2.t.t1+t2dt=โˆซโˆ’1โˆ’t2+11โˆ’t2dtI=tโˆ’โˆซ11โˆ’t2dtt=sinuโ†’dt=cosuduI=tโˆ’โˆซ1cos2ucosudu=1โˆ’x1+x+โˆซ1cosuduI=1โˆ’x1+x+โˆซcosu(1โˆ’sin2u)du=โ€ฆ..โˆ’12(โˆ’โˆซโˆ’cosu1โˆ’sinu+โˆซcosu1+sinudu)=โ€ฆ..โˆ’12[โˆ’ln(1โˆ’sinu)+ln(1+sinu)]=โ€ฆ..โˆ’12ln(1+sinu1โˆ’sinu)=โ€ฆ..โˆ’12ln(1+t1โˆ’t)=1โˆ’x1+xโˆ’12ln(1+1โˆ’x1+x1โˆ’1โˆ’x1+x)I=1โˆ’x1+xโˆ’12ln(1+x+1โˆ’x1+x1+xโˆ’1โˆ’x1+x)=1โˆ’x1+xโˆ’12ln(1+x+1โˆ’x1+xโˆ’1โˆ’x)=1โˆ’x1+xโˆ’12ln[(1+x+1โˆ’x)2(1+x)2โˆ’(1โˆ’x)2]=1โˆ’x1+xโˆ’12ln[1+1โˆ’x2x]
Commented by peter frank last updated on 28/Mar/22
thanks
thanks
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=((โˆš(1โˆ’x))/( (โˆš(1+x)))) โ‡” x=((1โˆ’t^2 )/(1+t^2 ))  dx=(dt/((d/dx)[((โˆš(1โˆ’x))/( (โˆš(1+x))))]))=(dt/(โˆ’(1/( (โˆš(1โˆ’x))(โˆš((1+x)^3 ))))))=โˆ’(โˆš(1โˆ’x))(โˆš((1+x)^3 ))dt  โ‡’  โˆซ((โˆš(1โˆ’x))/(x(โˆš(1+x))))dx=โˆซ((โˆš(1โˆ’x))/(x(โˆš(1+x))))ร—(โˆ’(โˆš(1โˆ’x))(โˆš((1+x)^3 ))dt)=  =โˆซ(((xโˆ’1)(x+1))/x)dt=โˆซ((((โˆ’2t^2 )/(t^2 +1))ร—(2/(t^2 +1)))/(โˆ’((t^2 โˆ’1)/(t^2 +1))))dt=  =4โˆซ(t^2 /((t^2 โˆ’1)(t^2 +1)))dt
thisiswrong.whatyoudointhe1stand3rdlinesmakesnosense.t=1โˆ’x1+xโ‡”x=1โˆ’t21+t2dx=dtddx[1โˆ’x1+x]=dtโˆ’11โˆ’x(1+x)3=โˆ’1โˆ’x(1+x)3dtโ‡’โˆซ1โˆ’xx1+xdx=โˆซ1โˆ’xx1+xร—(โˆ’1โˆ’x(1+x)3dt)==โˆซ(xโˆ’1)(x+1)xdt=โˆซโˆ’2t2t2+1ร—2t2+1โˆ’t2โˆ’1t2+1dt==4โˆซt2(t2โˆ’1)(t2+1)dt
Commented by ArielVyny last updated on 29/Mar/22
yes you have right. i have forgot tne square
yesyouhaveright.ihaveforgottnesquare
Answered by MJS_new last updated on 28/Mar/22
โˆซ((โˆš(1โˆ’x))/(x(โˆš(1+x))))dx=       [t=((โˆš(1โˆ’x))/( (โˆš(1+x)))) โ†’ dx=โˆ’(โˆš(1โˆ’x))(โˆš((1+x)^3 ))dt]  =4โˆซ(t^2 /(t^4 โˆ’1))dt=โˆซ((2/(t^2 +1))+(1/(tโˆ’1))โˆ’(1/(t+1)))dt=  =2arctan t +ln ((tโˆ’1)/(t+1)) =  =2arctan ((โˆš(1โˆ’x))/( (โˆš(1+x)))) +ln โˆฃ((1โˆ’(โˆš(1โˆ’x^2 )))/x)โˆฃ +C
โˆซ1โˆ’xx1+xdx=[t=1โˆ’x1+xโ†’dx=โˆ’1โˆ’x(1+x)3dt]=4โˆซt2t4โˆ’1dt=โˆซ(2t2+1+1tโˆ’1โˆ’1t+1)dt==2arctant+lntโˆ’1t+1==2arctan1โˆ’x1+x+lnโˆฃ1โˆ’1โˆ’x2xโˆฃ+C

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