∫_( (√2)) ^(3(√2)) (√(x^2 −2))dx+∫


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Question Number 139427 by Fikret last updated on 26/Apr/21

$\underset{\sqrt{2}}{\int^{3 \sqrt{2}}} \sqrt{x^{2} - 2} d x + \underset{0}{\int^{4}} \sqrt{x^{2} + 2} d x = ?$
	Answered by MJS_new last updated on 27/Apr/21

	$\int \sqrt{x^{2} - 2} d x =$ $[t = \frac{x + \sqrt{x^{2} - 2}}{\sqrt{2}} \to d x = \frac{\sqrt{2 (x^{2} - 2)}}{x + \sqrt{x^{2} - 2}} d t]$ $= \int (\frac{t}{2} - \frac{1}{t} + \frac{1}{2 t^{3}}) d t =$ $= \frac{t^{4} - 1}{4 t^{2}} - \ln t =$ $= \frac{1}{2} x \sqrt{x^{2} - 2} - \ln (x + \sqrt{x^{2} - 2}) + C_{1}$ $\int \sqrt{x^{2} + 2} d x =$ $[u = \frac{x + \sqrt{x^{2} + 2}}{\sqrt{2}} \to d x = \frac{\sqrt{2 (x^{2} + 2)}}{x + \sqrt{x^{2} + 2}} d u]$ $= \int (\frac{u}{2} + \frac{1}{u} + \frac{1}{2 u^{3}}) d t =$ $= \frac{u^{4} - 1}{4 u^{2}} + \ln u =$ $= \frac{1}{2} x \sqrt{x^{2} + 2} - \ln (x + \sqrt{x^{2} + 2}) + C_{2}$ $\Rightarrow answer is 12 \sqrt{2}$
	Answered by mathmax by abdo last updated on 27/Apr/21
	$I = H +K H =∫_(√2) ^(3(√2)) (√(x^2 −2))dx =_(x=(√2)cht→t=argch((x/( (√2))))) ∫_(argch(1)) ^(argch(3)) (√2)sht .(√2)sht dt =2 ∫_0 ^(ln(3+2(√2))) ((ch(2t)−1)/2)dt=[((sh(2t))/2)]_0 ^(ln(3+2(√2))) −ln(3+2(√2)) =(1/2)[((e^(2t) −e^(−2t) )/2)]_0 ^(ln(3+2(√2))) −ln(3+2(√2)) =(1/4){(3+2(√2))^2 −(3+2(√2))^(−2) } −ln(3+2(√2)) K =∫_0 ^4 (√(x^2 +2))dx =_(x=(√2)sh(t)→t=argsh((x/( (√2))))) ∫_0 ^(argsh((4/( (√2))))) (√2)ch(t)(√2)ch(t) =2 ∫_0 ^(ln((4/( (√2)))+3)) ((1+ch(2t))/2)dt =ln((4/( (√2)))+3)+[(1/2)sh(2t)]_0 ^(ln(3+2(√2))) =ln(3+2(√2))+(1/4){(3+2(√2))^2 −(3+2(√2))^(−2) } ⇒ I =(1/2){(3+2(√2))^2 −(3+2(√2))^(−2) }−ln(3+2(√2))$
	$I = H + K$ $H = \int_{\sqrt{2}}^{3 \sqrt{2}} \sqrt{x^{2} - 2} dx =_{x = \sqrt{2} cht \to t = argch (\frac{x}{\sqrt{2}})} \int_{argch (1)}^{argch (3)} \sqrt{2} sht . \sqrt{2} sht dt$ $= 2 \int_{0}^{\ln (3 + 2 \sqrt{2})} \frac{ch (2 t) - 1}{2} dt = {[\frac{sh (2 t)}{2}]}_{0}^{\ln (3 + 2 \sqrt{2})} - \ln (3 + 2 \sqrt{2})$ $= \frac{1}{2} {[\frac{e^{2 t} - e^{- 2 t}}{2}]}_{0}^{\ln (3 + 2 \sqrt{2})} - \ln (3 + 2 \sqrt{2}) = \frac{1}{4} {{(3 + 2 \sqrt{2})}^{2} - {(3 + 2 \sqrt{2})}^{- 2}}$ $- \ln (3 + 2 \sqrt{2})$ $K = \int_{0}^{4} \sqrt{x^{2} + 2} dx =_{x = \sqrt{2} sh (t) \to t = argsh (\frac{x}{\sqrt{2}})} \int_{0}^{argsh (\frac{4}{\sqrt{2}})} \sqrt{2} ch (t) \sqrt{2} ch (t)$ $= 2 \int_{0}^{\ln (\frac{4}{\sqrt{2}} + 3)} \frac{1 + ch (2 t)}{2} dt = \ln (\frac{4}{\sqrt{2}} + 3) + {[\frac{1}{2} sh (2 t)]}_{0}^{\ln (3 + 2 \sqrt{2})}$ $= \ln (3 + 2 \sqrt{2}) + \frac{1}{4} {{(3 + 2 \sqrt{2})}^{2} - {(3 + 2 \sqrt{2})}^{- 2}} \Rightarrow$ $I = \frac{1}{2} {{(3 + 2 \sqrt{2})}^{2} - {(3 + 2 \sqrt{2})}^{- 2}} - \ln (3 + 2 \sqrt{2})$
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