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Question Number 207620 by Ghisom last updated on 21/May/24
prove that ∫_(−a) ^a  (dx/(x^n +1+(√(x^(2n) +1))))=a
$$\mathrm{prove}\:\mathrm{that}\:\underset{−{a}} {\overset{{a}} {\int}}\:\frac{{dx}}{{x}^{{n}} +\mathrm{1}+\sqrt{{x}^{\mathrm{2}{n}} +\mathrm{1}}}={a} \\ $$
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
$$\int_{−{a}} ^{{a}} \frac{{x}^{{n}} +\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}{n}} }}{\mathrm{2}{x}^{{n}} }={U}_{{n}} \\ $$$$=\int_{−{a}} ^{{a}} \frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}{n}} }}{\mathrm{2}{x}^{{n}} }={a}+\int_{−{a}} ^{{a}} \frac{\mathrm{1}}{\mathrm{2}}+\int_{−{a}} ^{{a}} \frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}{n}} }}{\mathrm{2}{x}^{{n}} }{dx} \\ $$$$={a}+\frac{\mathrm{1}}{\mathrm{2}}\int_{−{a}} ^{{a}} \frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}{n}} }}{\mathrm{2}{x}^{{n}} }{dx} \\ $$$$\Rightarrow\forall{n}\in\mathbb{N}\int_{−{a}} ^{{a}} \frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}{n}} }}{\mathrm{2}{x}^{{n}} }=\mathrm{0}\:{you}\:{can}\:{see}\:{it}\:{if}\:\:{n}=\mathrm{2}{k} \\ $$$$\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}{n}} }}{\mathrm{2}{x}^{{n}} }\mathrm{0}\:\:<;\forall{x}\in\mathbb{R}−\left\{\mathrm{0}\right\} \\ $$$$\Rightarrow\int_{−{a}} ^{{a}} \frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}{n}} }}{\mathrm{2}{x}^{{n}} }<\mathrm{0}\neq\mathrm{0}\:{only}\:{if}\:{a}=\mathrm{0} \\ $$$${The}\:{resulta}\:{is}\:{true}\:{if}\:{n}=\mathrm{2}{k}+\mathrm{1};{k}\in\mathbb{N} \\ $$$$\overset{\ast} {{f}}=\begin{cases}{\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}{n}} }}{\mathrm{2}{x}^{{n}} }={x}\neq\mathrm{0};{n}=\mathrm{2}{k}+\mathrm{1};{k}\in\mathbb{N}}\\{=\mathrm{0}\:{if}\:{x}=\mathrm{0}}\end{cases} \\ $$$$\overset{\ast} {{f}}\left(−{x}\right)=−{f}\left({x}\right);\forall{x}\in\mathbb{R} \\ $$$$\int_{−{a}} ^{{a}} \overset{\ast} {{f}}\left({x}\right){dx}=\mathrm{0}\Rightarrow{U}_{\mathrm{2}{n}+\mathrm{1}} ={a} \\ $$$$ \\ $$
Commented by Ghisom last updated on 21/May/24
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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