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