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]](https://www.tinkutara.com/question/Q77242.png)
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...](https://www.tinkutara.com/question/Q77262.png)
Commented by Tony Lin last updated on 05/Jan/20

Commented by mathmax by abdo last updated on 05/Jan/20

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]](https://www.tinkutara.com/question/Q77265.png)
Commented by Tony Lin last updated on 05/Jan/20
