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Question-74720

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Question Number 74720 by chess1 last updated on 29/Nov/19
Commented by mathmax by abdo last updated on 29/Nov/19
changement (√((x−1)/(x+1))) =t give x−1 =t^2 (x+1) ⇒(1−t^2 )x=1+t^2 ⇒ x=((1+t^2 )/(1−t^2 )) ⇒(dx/dt) =((2t(1−t^2 )−(1+t^2 )(−2t))/((1−t^2 )^2 )) =((2t−2t^3 +2t +2t^3 )/((t^2 −1)^2 )) =((4t)/((t^2 −1)^2 )) ⇒∫ (√((x−1)/(x+1)))dx =4∫(t^2 /((t^2 −1)^2 ))dt let decompose F(t)=(t^2 /((t^2 −1)^2 )) ⇒F(t)=(t^2 /((t+1)^2 (t−1)^2 )) =(a/(t+1)) +(b/((t+1)^2 )) +(c/(t−1)) +(d/((t−1)^2 )) F(−t)=F(t) ⇒((−a)/(t−1)) +(b/((t−1)^2 )) −(c/(t+1)) +(d/((t+1)^2 )) =(a/(t+1)) +(b/((t+1)^2 )) +(c/(t−1)) +(d/((t−1)^2 )) ⇒c=−a and b=d ⇒ F(t)=(a/(t+1)) +(b/((t+1)^2 ))−(a/(t−1)) +(b/((t−1)^2 )) b=(t+1)^2 F(t)∣_(t=−1) =(1/4) ⇒F(t)=(a/(t+1)) +(1/(4(t+1)^2 ))−(a/(t−1)) +(1/(4(t−1)^2 )) F(0)=0 =a+(1/4) +a+(1/4) =2a+(1/2) ⇒2a=−(1/2) ⇒a=−(1/4) ⇒ F(t)=(1/(4(t−1)))−(1/(4(t+1))) +(1/(4(t+1)^2 )) +(1/(4(t−1)^2 )) ⇒ ∫(√((x−1)/(x+1)))dx =∫ (dt/(t−1))−∫ (dt/(t+1)) +∫ (dt/((t+1)^2 )) +∫ (dt/((t−1)^2 )) =ln∣t−1∣−ln∣t+1∣−(1/(t+1))−(1/(t−1)) +C =ln∣((t−1)/(t+1))∣−((1/(t+1))+(1/(t−1))) +C =ln∣((t−1)/(t+1))∣−((2t)/(t^2 −1)) +C =ln∣(((√((x−1)/(x+1 )))−1)/( (√((x−1)/(x+1)))+1))∣−((2(√((x−1)/(x+1))))/(((x−1)/(x+1))−1)) +C =ln∣(((√((x−1)/(x+1)))−1)/( (√((x−1)/(x+1)))+1))∣ +(x+1)(√((x−1)/(x+1 ))) + C .
changementx−1x+1=tgivex−1=t2(x+1)⇒(1−t2)x=1+t2⇒x=1+t21−t2⇒dxdt=2t(1−t2)−(1+t2)(−2t)(1−t2)2=2t−2t3+2t+2t3(t2−1)2=4t(t2−1)2⇒∫x−1x+1dx=4∫t2(t2−1)2dtletdecomposeF(t)=t2(t2−1)2⇒F(t)=t2(t+1)2(t−1)2=at+1+b(t+1)2+ct−1+d(t−1)2F(−t)=F(t)⇒−at−1+b(t−1)2−ct+1+d(t+1)2=at+1+b(t+1)2+ct−1+d(t−1)2⇒c=−aandb=d⇒F(t)=at+1+b(t+1)2−at−1+b(t−1)2b=(t+1)2F(t)∣t=−1=14⇒F(t)=at+1+14(t+1)2−at−1+14(t−1)2F(0)=0=a+14+a+14=2a+12⇒2a=−12⇒a=−14⇒F(t)=14(t−1)−14(t+1)+14(t+1)2+14(t−1)2⇒∫x−1x+1dx=∫dtt−1−∫dtt+1+∫dt(t+1)2+∫dt(t−1)2=ln∣t−1∣−ln∣t+1∣−1t+1−1t−1+C=ln∣t−1t+1∣−(1t+1+1t−1)+C=ln∣t−1t+1∣−2tt2−1+C=ln∣x−1x+1−1x−1x+1+1∣−2x−1x+1x−1x+1−1+C=ln∣x−1x+1−1x−1x+1+1∣+(x+1)x−1x+1+C.
Commented by mathmax by abdo last updated on 29/Nov/19
we can simplify we get ∫ (√((x−1)/(x+1)))dx =ln∣(((√(x−1))−(√(x+1)))/( (√(x−1)) +(√(x+1))))∣+(√(x^2 −1)) +C .
wecansimplifyweget∫x−1x+1dx=ln∣x−1−x+1x−1+x+1∣+x2−1+C.
Commented by chess1 last updated on 30/Nov/19
thanks
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Commented by mathmax by abdo last updated on 30/Nov/19
you are welcome.
youarewelcome.
Commented by mathmax by abdo last updated on 06/Dec/19
another way we use the changement x =ch(2t) ⇒ I =∫(√((ch(2t)−1)/(ch(2t)+1)))2sh(2t)dt =2∫ (√((2sh^2 (t))/(2ch^2 (t))))ch(2t)dt =2 ∫ th(t)2sh(t)ch(t)dt =4 ∫ ((sh(t))/(ch(t)))sh(t)ch(t)dt =4 ∫ sh^2 t dt =4 ∫((ch(2t)−1)/2)dt =2 ∫ ch(2t)dt−2t =sh(2t)−2t +c =(√(ch^2 (2t)−1)) −argch(x) +c =(√(x^2 −1)) −ln(x+(√(x^2 −1))) +c
anotherwayweusethechangementx=ch(2t)⇒I=∫ch(2t)−1ch(2t)+12sh(2t)dt=2∫2sh2(t)2ch2(t)ch(2t)dt=2∫th(t)2sh(t)ch(t)dt=4∫sh(t)ch(t)sh(t)ch(t)dt=4∫sh2tdt=4∫ch(2t)−12dt=2∫ch(2t)dt−2t=sh(2t)−2t+c=ch2(2t)−1−argch(x)+c=x2−1−ln(x+x2−1)+c
Answered by Tanmay chaudhury last updated on 29/Nov/19
∫((x−1)/( (√(x^2 −1))))dx (1/2)∫((d(x^2 −1))/((x^2 −1)^(1/2) ))−∫(dx/( (√(x^2 −1)))) I_1 =(1/2)×(((x^2 −1)^(((−1)/2)+1) )/(((−1)/2)+1))+c_1 ⇛(x^2 −1)^(1/2) +c_1 I_2 =∫(dx/( (√(x^2 −1)))) let x=secα dx=secαtanαdα =∫((secαtanαdα)/(tanα))=∫secαdα=ln(secα+tanα)+c_2 =ln(x+(√(x^2 −1)) )+c_2
∫x−1x2−1dx12∫d(x2−1)(x2−1)12−∫dxx2−1I1=12×(x2−1)−12+1−12+1+c1⇛(x2−1)12+c1I2=∫dxx2−1letx=secαdx=secαtanαdα=∫secαtanαdαtanα=∫secαdα=ln(secα+tanα)+c2=ln(x+x2−1)+c2
Commented by chess1 last updated on 30/Nov/19
thanks
thanks
Answered by Kunal12588 last updated on 30/Nov/19
I=∫(√((x−1)/(x+1)))dx=∫((x−1)/( (√(x^2 −1))))dx =∫(x/( (√(x^2 −1))))dx−∫(dx/( (√(x^2 −1)))) =(1/2)∫((2x)/( (√(x^2 −1))))dx−log∣x+(√(x^2 −1))∣+C let x^2 −1=t⇒2xdx=dt I=(1/2)∫(dt/( (√t)))−log∣x+(√(x^2 −1))∣+C =(1/2)(((√t)/(1/2)))−log∣x+(√(x^2 −1))∣+C =(√(x^2 −1))−log∣x+(√(x^2 −1))∣+C
I=∫x−1x+1dx=∫x−1x2−1dx=∫xx2−1dx−∫dxx2−1=12∫2xx2−1dx−log∣x+x2−1∣+Cletx2−1=t⇒2xdx=dtI=12∫dtt−log∣x+x2−1∣+C=12(t1/2)−log∣x+x2−1∣+C=x2−1−log∣x+x2−1∣+C

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