Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 146678 by mathdanisur last updated on 14/Jul/21

$$\int \sqrt{2+x^2}dx=?$$

Answered by Ar Brandon last updated on 14/Jul/21

$$\mathrm{I}=\int \sqrt{2+{\mathrm{x}}^2}\mathrm{dx}$$$$=2\int \sqrt{1+{\sinh }^2\theta }\cdot \cosh \theta \mathrm{d}\theta$$$$=2\int {\cosh }^2\theta =\int (\cosh 2\theta +1)\mathrm{d}\theta$$$$=\frac{\sinh 2\theta }{2}+\theta =\sinh \theta \cosh \theta +\theta +\mathrm{C}$$$$=\frac{\mathrm{x}}{2}\sqrt{2+{\mathrm{x}}^2}+\mathrm{argsinh}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}$$

Commented by mathdanisur last updated on 15/Jul/21

$$thankyouSercool$$

Answered by mathmax by abdo last updated on 14/Jul/21

$$\mathrm{I}=\int \sqrt{2+{\mathrm{x}}^2}\mathrm{dx}\mathrm{changement}\mathrm{x}=\sqrt{2}\mathrm{sht}\mathrm{give}$$$$\mathrm{I}=\int \sqrt{2}\mathrm{cht}\left(\sqrt{2}\right)\mathrm{cht}\mathrm{dt}=\int {2\mathrm{ch}}^2\mathrm{t}\mathrm{dt}=\int (1+\mathrm{ch}\left(2\mathrm{t}\right))\mathrm{dt}$$$$=\mathrm{t}+\frac{1}{2}\mathrm{sh}\left(2\mathrm{t}\right)=\mathrm{t}+\mathrm{sht}\mathrm{cht}\mathrm{but}\mathrm{t}=\mathrm{argsh}\left(\frac{\mathrm{x}}{\sqrt{2}}\right)=\log (\frac{\mathrm{x}}{\sqrt{2}}+\sqrt{1+\frac{{\mathrm{x}}^2}{2}})$$$$\Rightarrow \mathrm{I}=\log (\frac{\mathrm{x}}{\sqrt{2}}+\sqrt{1+\frac{{\mathrm{x}}^2}{2}})+\frac{\mathrm{x}}{\sqrt{2}}\sqrt{1+\frac{{\mathrm{x}}^2}{2}}+\mathrm{c}$$$$=\log (\mathrm{x}+\sqrt{2+{\mathrm{x}}^2})+\frac{\mathrm{x}}{2}\sqrt{2+{\mathrm{x}}^2}+\mathrm{C}$$

Commented by mathdanisur last updated on 15/Jul/21

$$thankyouSercool$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com