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x-1-x-dx-




Question Number 82566 by jagoll last updated on 22/Feb/20
∫ (√((x+1)/x)) dx = ?
x+1xdx=?
Commented by mathmax by abdo last updated on 22/Feb/20
let I=∫(√((x+1)/x))dx changement (√((x+1)/x))=t give ((x+1)/x)=t^2  ⇒  x+1 =xt^2  ⇒(1−t^2 )x=−1 ⇒x =(1/(t^2 −1)) ⇒dx =((−2tdt)/((t^2 −1)^2 )) ⇒  I =−2 ∫ t^2 ×(dt/((t^2 −1)^2 ))  =∫ ((((−2t))/((t^2 −1)^2 )))tdt by parts u^′  =((−2t)/((t^2 −1)^2 ))  and v=t ⇒ I =(t/(t^2 −1))−∫ (1/(t^2 −1))dt  =(t/(t^2 −1))−(1/2)∫((1/(t−1))−(1/(t+1)))dt  =(t/(t^2 −1))−(1/2)ln∣((t−1)/(t+1))∣ +c =((√((x+1)/x))/(((x+1)/x)−1))−(1/2)ln∣(((√((x+1)/x))−1)/( (√((x+1)/x))+1))∣ +C  =x(√((x+1)/x)) −(1/2)ln∣(((√((x+1)/x))−1)/( (√((x+1)/x))+1))∣ +C.
letI=x+1xdxchangementx+1x=tgivex+1x=t2x+1=xt2(1t2)x=1x=1t21dx=2tdt(t21)2I=2t2×dt(t21)2=((2t)(t21)2)tdtbypartsu=2t(t21)2andv=tI=tt211t21dt=tt2112(1t11t+1)dt=tt2112lnt1t+1+c=x+1xx+1x112lnx+1x1x+1x+1+C=xx+1x12lnx+1x1x+1x+1+C.
Answered by jagoll last updated on 22/Feb/20
Commented by mathmax by abdo last updated on 22/Feb/20
sir jagol (√((x+1)/x))≠((√(x+1))/( (√x)))
sirjagolx+1xx+1x
Commented by john santu last updated on 22/Feb/20
why (√((x+1)/x)) ≠ ((√(x+1))/( (√x)))   (√(2/3)) ≠ ((√2)/( (√3))) ??   i think it same sir
whyx+1xx+1x2323??ithinkitsamesir
Commented by mathmax by abdo last updated on 23/Feb/20
no sir (√((x+1)/x))is defined on]−∞,−1[∪]0,+∞[  but ((√(x+1))/( (√x)))  is defined on ]0,+∞[  (its not equals)
nosirx+1xisdefinedon],1[]0,+[butx+1xisdefinedon]0,+[(itsnotequals)
Answered by niroj last updated on 22/Feb/20
 Let I =∫(√(((x+1)^2 )/(x(x+1))))   dx  = ∫ ((x+1)/( (√(x(x+1)))))dx   = ∫ (x/( (√(x^2 +x))))dx+ ∫(( 1)/( (√(x^2 +x))))dx    I_1 =∫ (x/( (√(x^2 +x))))dx & I_2 =∫ (1/( (√(x^2 +x))))dx    For(I_1 ),put x^2 =t⇒ x=(√t)        2xdx=dt            xdx=(1/2)dt     I_1 = ∫ (1/(2(√(t+(√t)))))dt   I_1 = (1/2)∫ (( 1)/( (√(((√t) )^2 +2.(√t)(1/2)+(1/4)−(1/4)))))dt   I=I_1 +I_2       = (1/2)∫(1/( (√(((√t)  +(1/2))^2 −((1/2))^2 ))))dt +∫ (1/( (√((x)^2 +2.x.(1/2)+(1/4)−(1/4)))))dx  = (1/2)log( (√t)+(1/2)+ (√(((√t)  +(1/(2 )))^2 −(1/4))) )+∫(1/( (√((x+(1/2))^2 −(1/4)))))dx  = (1/2)log((√t)+(1/2)+(√(t+(√t)))  )+log(x+(1/2)+(√(x^2 +x)) )+C  = (1/2)log(x+(1/2)+(√(x^2 +x)) )+log(x+(1/2)+(√(x^2 +x)) )+C  = log(x+(1/2)+(√(x^2 +x)) )^(1/2) .(x+(1/2)+(√(x^2 +x)) )+C   = log(x+(1/2)+(√(x^2 +x))  )^(3/2) +C   = (3/2)log(((2x+1+2(√(x^2 +x)) )/2))+C   = (3/2)log(((2x+1+2(√(x(x+1))))/2))+c  = (3/2)log  ((((√x)  + (√(x+1)))^2 )/2)+c  = (3/2)[ log ((√x)  +(√(x+1)) )^2 −log2]+c  = (3/2).2log((√x)  +(√(x+1)) )−(3/2)log2+C  =  log ((√x)   +(√(x+1))  )^3 − (3/2)log2+C//.
LetI=(x+1)2x(x+1)dx=x+1x(x+1)dx=xx2+xdx+1x2+xdxI1=xx2+xdx&I2=1x2+xdxFor(I1),putx2=tx=t2xdx=dtxdx=12dtI1=12t+tdtI1=121(t)2+2.t12+1414dtI=I1+I2=121(t+12)2(12)2dt+1(x)2+2.x.12+1414dx=12log(t+12+(t+12)214)+1(x+12)214dx=12log(t+12+t+t)+log(x+12+x2+x)+C=12log(x+12+x2+x)+log(x+12+x2+x)+C=log(x+12+x2+x)12.(x+12+x2+x)+C=log(x+12+x2+x)32+C=32log(2x+1+2x2+x2)+C=32log(2x+1+2x(x+1)2)+c=32log(x+x+1)22+c=32[log(x+x+1)2log2]+c=32.2log(x+x+1)32log2+C=log(x+x+1)332log2+C//.

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