prove that ∫_(−a) ^a (dx/(x^n +1+(√(x^(2n) +1))))=a


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Question Number 207620 by Ghisom last updated on 21/May/24

$prove that \underset{- a}{\int^{a}} \frac{d x}{x^{n} + 1 + \sqrt{x^{2 n} + 1}} = a$
	Answered by Berbere last updated on 21/May/24
	$∫_(−a) ^a ((x^n +1−(√(1+x^(2n) )))/(2x^n ))=U_n =∫_(−a) ^a (1/2)+((1−(√(1+x^(2n) )))/(2x^n ))=a+∫_(−a) ^a (1/2)+∫_(−a) ^a ((1−(√(1+x^(2n) )))/(2x^n ))dx =a+(1/2)∫_(−a) ^a ((1−(√(1+x^(2n) )))/(2x^n ))dx ⇒∀n∈N∫_(−a) ^a ((1−(√(1+x^(2n) )))/(2x^n ))=0 you can see it if n=2k ((1−(√(1+x^(2n) )))/(2x^n ))0 <;∀x∈R−{0} ⇒∫_(−a) ^a ((1−(√(1+x^(2n) )))/(2x^n ))<0≠0 only if a=0 The resulta is true if n=2k+1;k∈N f^∗ = { ((((1−(√(1+x^(2n) )))/(2x^n ))=x≠0;n=2k+1;k∈N)),((=0 if x=0)) :} f^∗ (−x)=−f(x);∀x∈R ∫_(−a) ^a f^∗ (x)dx=0⇒U_(2n+1) =a$
	$\int_{- a}^{a} \frac{x^{n} + 1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} = U_{n}$ $= \int_{- a}^{a} \frac{1}{2} + \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} = a + \int_{- a}^{a} \frac{1}{2} + \int_{- a}^{a} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} d x$ $= a + \frac{1}{2} \int_{- a}^{a} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} d x$ $\Rightarrow \forall n \in N \int_{- a}^{a} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} = 0 y o u c a n s e e i t i f n = 2 k$ $\frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} 0 <; \forall x \in R - {0}$ $\Rightarrow \int_{- a}^{a} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} < 0 \neq 0 o n l y i f a = 0$ $T h e r e s u l t a i s t r u e i f n = 2 k + 1; k \in N$ $\overset{}{f} = {\begin{cases} \frac{1 - \sqrt{1 + x^{2 n}}}{2 x^{n}} = x \neq 0; n = 2 k + 1; k \in N \\ = 0 i f x = 0 \end{cases}$ $\overset{}{f} (- x) = - f (x); \forall x \in R$ $\int_{- a}^{a} \overset{*}{f} (x) d x = 0 \Rightarrow U_{2 n + 1} = a$
	Commented by Ghisom last updated on 21/May/24

	$thank you$
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