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Question Number 144402 by rexford last updated on 25/Jun/21

Commented by justtry last updated on 25/Jun/21

Commented by rexford last updated on 27/Jun/21

$$thanksverymuchforyourtime$$$$Iamverygrateful$$$$$$

Answered by mathmax by abdo last updated on 25/Jun/21

$$\mathrm{\Psi }=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{{27\mathrm{x}}^2+6\mathrm{x}-2}}\Rightarrow$$$${\mathrm{\Delta }}^{{}^{\prime }}=3^2-27.(-2)=9+54=63\Rightarrow {\mathrm{x}}_1=\frac{-3+\sqrt{63}}{27}$$$${\mathrm{x}}_2=\frac{-3-\sqrt{63}}{27}\Rightarrow \mathrm{\Psi }=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{27(\mathrm{x}-{\mathrm{x}}_1)(\mathrm{x}-{\mathrm{x}}_2)}}$$$$=\frac{1}{\sqrt{27}}\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{\mathrm{x}-{\mathrm{x}}_1}\sqrt{\mathrm{x}-{\mathrm{x}}_2}}$$$${=}_{\sqrt{\mathrm{x}-{\mathrm{x}}_1}=\mathrm{t}}\frac{1}{\sqrt{27}}\int \frac{2\mathrm{tdt}}{{({\mathrm{t}}^2+{\mathrm{x}}_1)}^2\mathrm{t}\sqrt{{\mathrm{t}}^2+{\mathrm{x}}_1-{\mathrm{x}}_2}}$$$$=\frac{2}{\sqrt{27}}\int \frac{\mathrm{dt}}{{({\mathrm{t}}^2+{\mathrm{x}}_1)}^2\sqrt{{\mathrm{t}}^2+{\mathrm{x}}_1-{\mathrm{x}}_2}}$$$$\mathrm{we}\mathrm{have}{\mathrm{x}}_1-{\mathrm{x}}_2=\frac{2\sqrt{63}}{27}\mathrm{so}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{cha}7\mathrm{gement}\mathrm{t}=\sqrt{\frac{2\sqrt{63}}{27}}\mathrm{shy}$$$$\Rightarrow \mathrm{\Psi }=\frac{2}{\sqrt{27}}\int \frac{{\lambda }_0\mathrm{chy}\mathrm{dy}}{{({\lambda }_0^2{\mathrm{sh}}^2\mathrm{y}+{\mathrm{x}}_1)}^2{\lambda }_0\mathrm{chy}}$$$$=\frac{2}{\sqrt{27}}\int \frac{\mathrm{dy}}{{({\lambda }_0^2{\mathrm{sh}}^2\mathrm{y}+{\mathrm{x}}_1)}^2}({\lambda }_0=\sqrt{\frac{2\sqrt{63}}{27}})$$$$....\mathrm{be}\mathrm{continued}...$$

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