Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 157455 by aliibrahim1 last updated on 23/Oct/21

Answered by FongXD last updated on 23/Oct/21

$$\mathrm{let}\mathrm{u}={\mathrm{x}}^{2021}+\frac{1}{{\mathrm{x}}^{2021}+{.}_{{.}_{{.}_{.}}}}={\mathrm{x}}^{2021}+\frac{1}{\mathrm{u}}$$$$\iff {\mathrm{u}}^2-{\mathrm{x}}^{2021}\mathrm{u}-1=0,\Rightarrow \mathrm{u}=\frac{{\mathrm{x}}^{2021}+\sqrt{{\mathrm{x}}^{4042}+4}}{2}$$$$\mathrm{and}\mathrm{v}={\mathrm{x}}^{2020}+\frac{1}{{\mathrm{x}}^{2020}+{.}_{{.}_{{.}_{.}}}}={\mathrm{x}}^{2020}+\frac{1}{\mathrm{v}}$$$$\iff {\mathrm{v}}^2-{\mathrm{x}}^{2020}\mathrm{v}-1=0,\Rightarrow \mathrm{v}=\frac{{\mathrm{x}}^{2020}+\sqrt{{\mathrm{x}}^{4040}+4}}{2}$$$$\mathrm{we}\mathrm{get}:\mathrm{I}=2{\int }_1^2(\frac{{\mathrm{x}}^{4041}}{{\mathrm{x}}^{2021}+\sqrt{{\mathrm{x}}^{4042}+4}}+\frac{{\mathrm{x}}^{4039}}{{\mathrm{x}}^{2020}+\sqrt{{\mathrm{x}}^{4040}+4}})\mathrm{dx}$$$$\iff \mathrm{I}=-\frac{1}{2}{\int }_1^2[{\mathrm{x}}^{4041}({\mathrm{x}}^{2021}-\sqrt{{\mathrm{x}}^{4042}+4})+{\mathrm{x}}^{4039}({\mathrm{x}}^{2020}-\sqrt{{\mathrm{x}}^{4040}+4})]\mathrm{dx}$$$$\iff \mathrm{I}=-\frac{1}{2}{\int }_1^2({\mathrm{x}}^{6062}+{\mathrm{x}}^{6059})\mathrm{dx}+\frac{1}{2}{\int }_1^2\frac{1}{4042}{({\mathrm{x}}^{4042}+4)}^{\frac{1}{2}}{({\mathrm{x}}^{4042}+4)}^{\prime }\mathrm{dx}+\frac{1}{2}{\int }_1^2\frac{1}{4040}{({\mathrm{x}}^{4040}+4)}^{\frac{1}{2}}{({\mathrm{x}}^{4040}+4)}^{\prime }\mathrm{dx}$$$$\iff \mathrm{I}=-\frac{1}{2}{[\frac{{\mathrm{x}}^{6063}}{6063}+\frac{{\mathrm{x}}^{6060}}{6060}-\frac{2{({\mathrm{x}}^{4042}+4)}^{\frac{3}{2}}}{3\times 4042}-\frac{2{({\mathrm{x}}^{4040}+4)}^{\frac{3}{2}}}{3\times 4040}]}_1^2$$$$\mathrm{Continue}...$$

Commented by aliibrahim1 last updated on 24/Oct/21

$$thxsir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com