prove that ∫_0 ^a (√(2+(a/x)−2(√(a/x)) ))dx=a[(1/(√2))ln((√2)+1)+1]


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Question Number 77242 by Tony Lin last updated on 04/Jan/20

$p r o v e t h a t$ $\int_{0}^{a} \sqrt{2 + \frac{a}{x} - 2 \sqrt{\frac{a}{x}}} d x = a [\frac{1}{\sqrt{2}} l n (\sqrt{2} + 1) + 1]$
Commented by mathmax by abdo last updated on 04/Jan/20

$l e t f (a) = \int_{0}^{a} \sqrt{2 + \frac{a}{x} - 2 \sqrt{\frac{a}{x}}} d x c h a n g e m e n t \sqrt{\frac{a}{x}} = t h i v e$ $\frac{a}{x} = t^{2} \Rightarrow a = {x t}^{2} \Rightarrow x = {a t}^{- 2} \Rightarrow d x = - 2 {a t}^{- 3} d t a n d$ $f (a) = \int_{0}^{1} \sqrt{2 + t^{2} - 2 t} (- 2 a) t^{- 3} d t$ $= (- 2 a) \int_{0}^{1} \sqrt{t^{2} - 2 t + 1 + 1} t^{- 3} d t = - 2 a \int_{0}^{1} \sqrt{{(t - 1)}^{2} + 1} t^{- 3} d t$ $=_{t - 1 = s h (u)} - 2 a \int_{a r g s h (- 1)}^{0} c h (u) {(1 + s h (u))}^{- 3} c h (u) d u$ $= - 2 a \int_{- l n (1 + \sqrt{2})}^{0} {c h}^{2} u {(1 + s h (u))}^{2} d u$ $= - 2 a \int_{- l n (1 + \sqrt{2})}^{0} {c h}^{2} u (1 + 2 s h u + {s h}^{2} u) d u$ $= - 2 a \int_{- l n (1 + \sqrt{2})}^{0} {c h}^{2} u d u - 4 a \int_{- l n (1 + \sqrt{2})}^{0} s h u {c h}^{2} u d u$ $- 2 a \int_{- l n (1 + \sqrt{2})}^{0} {s h}^{2} u {c h}^{2} u d u$ $= - 2 a \int_{- l n (1 + \sqrt{2})}^{0} \frac{c h (2 u) + 1}{2} d u - 4 a {[\frac{1}{3} {c h}^{3} u]}_{- l n (1 + \sqrt{2})}^{0}$ $- \frac{a}{2} \int_{- l n (1 + \sqrt{2})}^{0} {s h}^{2} u d u$ $= a {[u + \frac{1}{2} s h u]}_{- l n (1 + \sqrt{2})}^{0} - \frac{4 a}{3} {[\frac{e^{3 u} + e^{- 3 u}}{2}]}_{- l n (1 + \sqrt{2})}^{0}$ $- \frac{a}{2} \int_{- l n (1 + \sqrt{2})}^{0} \frac{c h (2 u) - 1}{2} d u r e s t t o a c h i e v e t h e c a l c u l u s . . .$
Commented by Tony Lin last updated on 05/Jan/20

$t h a n k s s i r$
Commented by mathmax by abdo last updated on 05/Jan/20

$y o u a r e w e l c o m e .$
	Answered by MJS last updated on 05/Jan/20

	$a ⩾ 0 \land x > 0 \lor a ⩽ 0 \land x < 0$ $\Rightarrow$ $\int \sqrt{2 + \frac{a}{x} - 2 \sqrt{\frac{a}{x}}} d x = \int \frac{\sqrt{2 x - 2 \sqrt{a x} + a}}{\sqrt{x}} d x =$ $[t = \sqrt{x} \to d x = 2 \sqrt{x} d t]$ $= 2 \int \sqrt{t^{2} - 2 \sqrt{a} t + a} d t =$ $= \frac{2 t - \sqrt{a}}{2} \sqrt{2 t^{2} - 2 \sqrt{a} t + a} + \frac{\sqrt{2}}{4} a \ln (2 t - \sqrt{a} + \sqrt{4 t^{2} - 4 \sqrt{a} t + 2 a}) =$ $= \frac{2 \sqrt{x} - \sqrt{a}}{2} \sqrt{2 x - 2 \sqrt{a x} + a} + \frac{\sqrt{2}}{4} a \ln (2 \sqrt{x} - \sqrt{a} + \sqrt{4 x - 4 \sqrt{a x} + 2 a}) + C$ $now just insert the values x = [0; a]$
	Commented by Tony Lin last updated on 05/Jan/20

	$t h a n k s s i r$
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