Menu Close

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=∫(1/x)((√((1−x)/(1+x))))dx  Indication poser t=(√((1−x)/(1+x)))
CalculateI=1x(1x1+x)dxIndicationposert=1x1+x
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)]
t=1x1+xdt=2(1+x)22tdx=1(1+x)21tdxt2=1x1+x1t2=x(t2+1)x=1t21+t2dt=1(1+t21+t2+1t21+t2)tdx=.1+t2tdxI=1+t21t2.t.t1+t2dt=1t2+11t2dtI=t11t2dtt=sinudt=cosuduI=t1cos2ucosudu=1x1+x+1cosuduI=1x1+x+cosu(1sin2u)du=..12(cosu1sinu+cosu1+sinudu)=..12[ln(1sinu)+ln(1+sinu)]=..12ln(1+sinu1sinu)=..12ln(1+t1t)=1x1+x12ln(1+1x1+x11x1+x)I=1x1+x12ln(1+x+1x1+x1+x1x1+x)=1x1+x12ln(1+x+1x1+x1x)=1x1+x12ln[(1+x+1x)2(1+x)2(1x)2]=1x1+x12ln[1+1x2x]
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=((√(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
thisiswrong.whatyoudointhe1stand3rdlinesmakesnosense.t=1x1+xx=1t21+t2dx=dtddx[1x1+x]=dt11x(1+x)3=1x(1+x)3dt1xx1+xdx=1xx1+x×(1x(1+x)3dt)==(x1)(x+1)xdt=2t2t2+1×2t2+1t21t2+1dt==4t2(t21)(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
∫((√(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
1xx1+xdx=[t=1x1+xdx=1x(1+x)3dt]=4t2t41dt=(2t2+1+1t11t+1)dt==2arctant+lnt1t+1==2arctan1x1+x+ln11x2x+C

Leave a Reply

Your email address will not be published. Required fields are marked *