Question Number 102128 by Study last updated on 06/Jul/20

 ∫((x+1)/((√x) +1))dx=?

Answered by Dwaipayan Shikari last updated on 06/Jul/20

∫(x/((√x)+1))+(1/((√x)+1))=∫((2u^3 )/(u+1))du+∫((2u)/(u+1))du    {   take x  as u^2   =2∫((u^3 +1)/(u+1))−(1/(u+1))du+2∫1−(1/(u+1))du  =2∫u^2 −u+1−2log(u+1)+2u−2log(u+1)  =2∫(u−1)^2 +u du+2u−4log(u+1)  =2∫(u−1)^2 du+u^2 +2u−4log(u+1)  =(2/3)(u−1)^3 +u^2 +2u−4log(u+1)+C  =(2/3)((√x)−1)^3 +x+2(√x)−4log((√x)+1)+C

Answered by abdomathmax last updated on 06/Jul/20

I =∫ ((x+1)/((√x)+1))dx changement (√x)=t give x =t^2  ⇒  I =∫ ((t^2  +1)/(t+1))(2t)dt =2 ∫ ((t^3  +t)/(t+1)) dt  =2 ∫  (((t+1−1)(t^2 )+t)/(t+1))dt  =2 ∫ t^2  dt +2 ∫((−t^2  +t)/(t+1)) dt  =(2/3)t^3  −2 ∫((t^2 −t)/(t+1))dt but  ∫ ((t^2 −t)/(t+1))dt =∫ ((t(t+1−1)−t)/(t+1))dt  =∫ t dt−2 ∫ (t/(t+1))dt  =(t^2 /2)−2 ∫ ((t+1−1)/(t+1))dt =(t^2 /2)−2t +2ln∣t+1∣ +c ⇒  I =(2/3)t^3 −t^2 +4t−4ln∣t+1∣ +C  I=(2/3)((√x))^(3 ) −x +4(√x)−4ln∣(√x)+1∣ +C

Commented bymathmax by abdo last updated on 06/Jul/20

do you want to get another answer...!

Commented byDwaipayan Shikari last updated on 07/Jul/20

Sorry sir!

Answered by Ar Brandon last updated on 06/Jul/20

I=∫((x+1)/((√x)+1))dx  ,  Let t=(√x)  ⇒((x+1)/((√x)+1))=((t^2 +1)/(t+1))=t−1+(2/(t+1))  ⇒I=∫{(√x)−1+(2/((√x)+1))}dx=∫{x^(1/2) −1+((2((√x)−1)/(x−1))}dx          =∫{x^(1/2) −1−(2/(x−1))}dx+2∫((√x)/(x−1))dx , u^2 =x ⇒2udu=dx          =((2x^(3/2) )/3)−x−2ln∣x−1∣+2∫(u/(u^2 −1))∙2udu          =((2x^(3/2) )/3)−x−2ln∣x−1∣+4∫(u^2 /(u^2 −1))du  K=∫(u^2 /(u^2 −1))du=∫{1+(1/(u^2 −1))}du=u−tanh^(−1) (u)  ⇒I=((2x^(3/2) )/3)−x−2ln∣x−1∣+(√x)−tanh^(−1) ((√x))          =((2(√x^3 ))/(√3))+(√x)−x−2ln∣x−1∣−tanh^(−1) ((√x))+C

Answered by OlafThorendsen last updated on 07/Jul/20

u = (√x)+1 ⇒ dx = 2(u−1)du  ∫(((u−1)^2 +1)/u)2(u−1)du  ∫2((u^2 −2u+2)/u)(u−1)du  ∫2(u^2 −2u+2)(1−(1/u))du  ∫2(u^2 −2u+2−u+2−(2/u))du  ∫2(u^2 −3u+4−(2/u))du  2((u^3 /3)−(3/2)u^2 +4u−2ln∣u∣)+C  (2/3)((√x)+1)^3 −3((√x)+1)^2 +8((√x)+1)−4ln((√x)+1)+C