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 171565 by Tawa11 last updated on 17/Jun/22

Commented by infinityaction last updated on 17/Jun/22

$$p=\frac{x+1}{x-1}then\frac{1}{p}=\frac{x-1}{x+1}$$$${(p^2+\frac{1}{p^2}-2)}^{1/2}={(p-\frac{1}{p})}^2$$$${\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{\left\{{(\frac{x-1}{x+1}-\frac{x+1}{x-1})}^2\right\}}^{1/2}dx$$$${\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{\left\{{\left(\frac{{\mathit{x}}^2-2\mathit{x}+1-{\mathit{x}}^2-2\mathit{x}-1}{{\mathit{x}}^2-1}\right)}^2\right\}}^{1/2}\mathit{d}\mathit{x}$$$${\int }_{-\frac{1}{\sqrt{2}}}^{1/\sqrt{2}}{\left\{{\left(\frac{4\mathit{x}}{1-x^2}\right)}^2\right\}}^{1/2}\mathit{d}\mathit{x}$$$${\int }_{-\frac{1}{\sqrt{2}}}^{1/\sqrt{2}}\mid \frac{4x}{1-x^2}\mid dx$$$${\int }_{-\frac{1}{\sqrt{2}}}^0-\frac{4\mathit{x}}{1-{\mathit{x}}^2}\mathit{d}\mathit{x}+{\int }_0^{1/\sqrt{2}}\frac{4\mathit{x}}{1-{\mathit{x}}^2}\mathit{d}\mathit{x}$$$$nowcomplete$$

Commented by immortels last updated on 18/Jun/22

$$il{y}^{\prime }auneerreur$$$$$$

Commented by infinityaction last updated on 18/Jun/22

$$where$$

Commented by immortels last updated on 19/Jun/22

$${(p^2+\frac{1}{p^2}-2)}^{1/2}={(p-\frac{1}{p})}^{2}icicadvientjuste$$$$(p-\frac{1}{p})verifiemaisapresc.{,}^{\prime }estbon$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com