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Question Number 128853 by Ar Brandon last updated on 10/Jan/21

$$\mathrm{F}\left(\mathrm{x}\right)={\int }_{\mathrm{x}}^{2\mathrm{x}}\frac{\mathrm{dx}}{\sqrt{{\mathrm{t}}^4+{\mathrm{t}}^2+1}}$$$$1.\mathrm{Show}\mathrm{that}\mathrm{F}\mathrm{is}\mathrm{defined},\mathrm{continuous}\mathrm{and}\mathrm{derivable}\mathrm{in}\mathbb{R}$$

Answered by mathmax by abdo last updated on 11/Jan/21

$$\mathrm{F}\left(\mathrm{x}\right)={\int }_{\mathrm{R}}\frac{1}{\sqrt{{\mathrm{t}}^4+{\mathrm{t}}^2+1}}{\chi }_{[\mathrm{x},2\mathrm{x}]}\left(\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{the}\mathrm{function}\mathrm{x}\to \frac{{\chi }_{[\mathrm{x},2\mathrm{x}]}\left(\mathrm{t}\right)}{\sqrt{{\mathrm{t}}^4+{\mathrm{t}}^2+1}}\mathrm{is}\mathrm{continue}\mathrm{and}\mathrm{derivable}\mathrm{so}$$$$\mathrm{F}\mathrm{is}\mathrm{derivable}\mathrm{and}{\mathrm{F}}^{{}^{\prime }}\left(\mathrm{x}\right)={\int }_{\mathrm{R}}\frac{\partial }{\partial \mathrm{x}}\left\{\frac{{\chi }_{[\mathrm{x},2\mathrm{x}]}\left(\mathrm{t}\right)}{\sqrt{{\mathrm{t}}^4+{\mathrm{t}}^2+1}}\right\}\mathrm{dt}$$$$=\frac{2}{\sqrt{{\left(2\mathrm{x}\right)}^4+{\left(2\mathrm{x}\right)}^2+1}}-\frac{1}{\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1}}=\frac{2}{\sqrt{{16\mathrm{x}}^4+{4\mathrm{x}}^2+1}}-\frac{1}{\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1}}$$

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