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Question Number 77242 by Tony Lin last updated on 04/Jan/20
prove that   ∫_0 ^a (√(2+(a/x)−2(√(a/x)) ))dx=a[(1/( (√2)))ln((√2)+1)+1]
provethat0a2+ax2axdx=a[12ln(2+1)+1]
Commented by mathmax by abdo last updated on 04/Jan/20
let f(a)=∫_0 ^a (√(2+(a/x)−2(√(a/x))))dx  changement (√(a/x))=t hive  (a/x) =t^2  ⇒a=xt^2  ⇒x=at^(−2)  ⇒dx =−2at^(−3)  dt and  f(a)=∫_0 ^1 (√(2+t^2 −2t))(−2a)t^(−3 ) dt  =(−2a)∫_0 ^1 (√(t^2 −2t+1+1))t^(−3)  dt =−2a∫_0 ^1 (√((t−1)^2 +1))t^(−3)  dt  =_(t−1 =sh(u))   −2a ∫_(argsh(−1)) ^0 ch(u)(1+sh(u))^(−3)  ch(u)du  =−2a ∫_(−ln(1+(√2))) ^0  ch^2 u(1+sh(u))^2  du  =−2a ∫_(−ln(1+(√2))) ^0  ch^2 u(1+2shu +sh^2 u)du  =−2a∫_(−ln(1+(√2))) ^0  ch^2 u du −4a ∫_(−ln(1+(√2))) ^0  shu ch^2 udu  −2a∫_(−ln(1+(√2))) ^0  sh^2 u ch^2 u du  =−2a∫_(−ln(1+(√2))) ^0 ((ch(2u)+1)/2)du −4a [(1/3)ch^3 u]_(−ln(1+(√2))) ^0   −(a/2) ∫_(−ln(1+(√2))) ^0  sh^2 u du   =a [u+(1/2)shu]_(−ln(1+(√2))) ^0 −((4a)/3)[((e^(3u) +e^(−3u) )/2)]_(−ln(1+(√2))) ^0   −(a/2)∫_(−ln(1+(√2))) ^0  ((ch(2u)−1)/2)du  rest to achieve the calculus...
letf(a)=0a2+ax2axdxchangementax=thiveax=t2a=xt2x=at2dx=2at3dtandf(a)=012+t22t(2a)t3dt=(2a)01t22t+1+1t3dt=2a01(t1)2+1t3dt=t1=sh(u)2aargsh(1)0ch(u)(1+sh(u))3ch(u)du=2aln(1+2)0ch2u(1+sh(u))2du=2aln(1+2)0ch2u(1+2shu+sh2u)du=2aln(1+2)0ch2udu4aln(1+2)0shuch2udu2aln(1+2)0sh2uch2udu=2aln(1+2)0ch(2u)+12du4a[13ch3u]ln(1+2)0a2ln(1+2)0sh2udu=a[u+12shu]ln(1+2)04a3[e3u+e3u2]ln(1+2)0a2ln(1+2)0ch(2u)12duresttoachievethecalculus
Commented by Tony Lin last updated on 05/Jan/20
thanks sir
thankssir
Commented by mathmax by abdo last updated on 05/Jan/20
you are welcome.
youarewelcome.
Answered by MJS last updated on 05/Jan/20
a≥0∧x>0 ∨ a≤0∧x<0  ⇒  ∫(√(2+(a/x)−2(√(a/x))))dx=∫((√(2x−2(√(ax))+a))/( (√x)))dx=       [t=(√x) → dx=2(√x)dt]  =2∫(√(t^2 −2(√a)t+a))dt=  =((2t−(√a))/2)(√(2t^2 −2(√a)t+a))+((√2)/4)aln (2t−(√a)+(√(4t^2 −4(√a)t+2a))) =  =((2(√x)−(√a))/2)(√(2x−2(√(ax))+a))+((√2)/4)aln (2(√x)−(√a)+(√(4x−4(√(ax))+2a))) +C  now just insert the values x=[0; a]
a0x>0a0x<02+ax2axdx=2x2ax+axdx=[t=xdx=2xdt]=2t22at+adt==2ta22t22at+a+24aln(2ta+4t24at+2a)==2xa22x2ax+a+24aln(2xa+4x4ax+2a)+Cnowjustinsertthevaluesx=[0;a]
Commented by Tony Lin last updated on 05/Jan/20
thanks sir
thankssir

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