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Question Number 128851 by mnjuly1970 last updated on 10/Jan/21
$...nice calculus (I)... calculate :: Ψ=∫_0 ^( 1) {(e−1)(√(log( 1+ex−x ))) +e^x^2 }dx=?$
$. . . n i c e c a l c u l u s (I) . . .$ $c a l c u l a t e ::$ $Ψ = \int_{0}^{1} {(e - 1) \sqrt{l o g (1 + e x - x)} + e^{x^{2}}} d x = ?$
Commented by mindispower last updated on 10/Jan/21
${.} fraction part ?$
${.} f r a c t i o n p a r t ?$
Commented by mnjuly1970 last updated on 11/Jan/21

$n o m r p o w e r$
	Answered by Olaf last updated on 11/Jan/21
	$Let f(x) = (e−1)(√(ln(1+(e−1)x))+e^x^2 f is clearly an increasing function and f(0) = 1, f(1) = e ⇒ ∃a∈]0,1[ \ f(a) = 2 Ψ = ∫_0 ^1 {f(x)}dx = ∫_0 ^a {f(x)}dx+∫_a ^1 {f(x)}dx Ψ = ∫_0 ^a (f(x)−1)dx+∫_a ^1 (f(x)−2)dx Ψ = ∫_0 ^1 f(x)dx−a−2(1−a) Ψ = ∫_0 ^1 f(x)dx+a−2 Let u = ln(1+(e−1)x) e^u = 1+(e−1)x e^u du = (e−1)dx I = (e−1)∫_0 ^1 (√(ln(1+(e−1)x)))dx I = (e−1)∫_0 ^1 (√u)((e^u du)/(e−1)) I = ∫_0 ^1 (√u)e^u du I = [(√x)e^x −((√π)/2)erfi((√x))]_0 ^1 = e−((√π)/2)erfi(1) J = ∫_0 ^1 e^x^2 dx = [((√π)/2)erfi(x)]_0 ^1 = ((√π)/2)erfi(1) Ψ = I+J+a−2 = e+a−2 (a ≈ 0,542)$
	$Let f (x) = (e - 1) \sqrt{\ln (1 + (e - 1) x} + e^{x^{2}}$ $f is clearly an increasing function and$ $f (0) = 1, f (1) = e$ $\Rightarrow \exists a \in] 0, 1 [∖ f (a) = 2$ $Ψ = \int_{0}^{1} {f (x)} d x = \int_{0}^{a} {f (x)} d x + \int_{a}^{1} {f (x)} d x$ $Ψ = \int_{0}^{a} (f (x) - 1) d x + \int_{a}^{1} (f (x) - 2) d x$ $Ψ = \int_{0}^{1} f (x) d x - a - 2 (1 - a)$ $Ψ = \int_{0}^{1} f (x) d x + a - 2$ $Let u = \ln (1 + (e - 1) x)$ $e^{u} = 1 + (e - 1) x$ $e^{u} d u = (e - 1) d x$ $I = (e - 1) \int_{0}^{1} \sqrt{\ln (1 + (e - 1) x)} d x$ $I = (e - 1) \int_{0}^{1} \sqrt{u} \frac{e^{u} d u}{e - 1}$ $I = \int_{0}^{1} \sqrt{u} e^{u} d u$ $I = {[\sqrt{x} e^{x} - \frac{\sqrt{π}}{2} erfi (\sqrt{x})]}_{0}^{1} = e - \frac{\sqrt{π}}{2} erfi (1)$ $J = \int_{0}^{1} e^{x^{2}} d x = {[\frac{\sqrt{π}}{2} erfi (x)]}_{0}^{1} = \frac{\sqrt{π}}{2} erfi (1)$ $Ψ = I + J + a - 2 = e + a - 2$ $(a \approx 0, 542)$
	Commented by mnjuly1970 last updated on 11/Jan/21

	$g r a t e f u l f o r y o u r w o r k m r o l a f . .$
	Answered by mnjuly1970 last updated on 11/Jan/21

	$n o t e : \int_{a}^{b} f (x) d x + \int_{c}^{d} f^{- 1} (x) d x = b d - a c$ $Ψ = \int_{0}^{1} (e - 1) \sqrt{l o g (1 + x (e - 1)} d x$ $+ \int_{0}^{1} e^{x^{2}} d x$ $∴ Ψ \overset{[u = (1 + x (e - 1)]}{=} \int_{1}^{e} \sqrt{l o g (u)} d u + \int_{0}^{1} e^{x^{2}} d x$ $= \int_{1}^{e} \sqrt{l o g (x)} + \int_{0}^{1} e^{x^{2}} d x = e$ $n o t e : f (x) = \sqrt{l o g (x)} \Leftrightarrow f^{- 1} (x) = e^{x^{2}}$
	Answered by mnjuly1970 last updated on 11/Jan/21

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