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 44476 by abdo.msup.com last updated on 29/Sep/18

$$letf\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+2xt-1}$$$$1)findaexplicitformoff\left(x\right)$$$$2)cslvulste{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+t-1}$$$$3)calculateA\left(\theta \right)={\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+2tan\left(\theta \right)t-1}$$$$4)calculateg\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{tdt}{{(t^2+2xt-1)}^2}$$$$5)findthevalueof{\int }_0^{\mathrm{\infty }}\frac{tdt}{{(t^2+4t-1)}^2}$$

Commented by maxmathsup by imad last updated on 01/Oct/18

$$1)wehavef\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+2xt+x^2-x^2-1}$$$$={\int }_0^{\mathrm{\infty }}\frac{dt}{{(t+x)}^2-(x^2+1)}={\int }_0^{\mathrm{\infty }}\frac{dt}{(t+x+\sqrt{1+x^2})(t+x-\sqrt{1+x^2})}$$$$=\frac{1}{2\sqrt{1+x^2}}{\int }_0^{\mathrm{\infty }}(\frac{1}{t+x-\sqrt{1+x^2}}-\frac{1}{t+x+\sqrt{1+x^2}})dt$$$$=\frac{1}{2\sqrt{1+x^2}}{[ln\mid \frac{t+x-\sqrt{1+x^2}}{t+x+\sqrt{1+x^2}}]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{2\sqrt{1+x^2}}(-ln\mid \frac{x-\sqrt{1+x^2}}{x+\sqrt{1+x^2}}\mid )$$$$\Rightarrow f\left(x\right)=\frac{1}{2\sqrt{1+x^2}}ln\mid \frac{x+\sqrt{1+x^2}}{x-\sqrt{1+x^2}}\mid .$$$$$$

Commented by maxmathsup by imad last updated on 01/Oct/18

$$2){\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+t-1}=f\left(\frac{1}{2}\right)=\frac{1}{2\sqrt{1+\frac{1}{4}}}ln\mid \frac{\frac{1}{2}+\sqrt{1+\frac{1}{4}}}{\frac{1}{2}-\sqrt{1+\frac{1}{4}}}\mid$$$$=\frac{1}{\sqrt{5}}ln\left(\frac{1+\sqrt{5}}{\sqrt{5}-1}\right).$$

Commented by maxmathsup by imad last updated on 01/Oct/18

$$3)wehave{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2-2tan\theta t-1}=f\left(tan\theta \right)$$$$=\frac{1}{2\sqrt{1+{tan}^2\theta }}ln\mid \frac{tan\theta +\sqrt{1+{tan}^2\theta }}{tan\theta -\sqrt{1+{tan}^2\theta }}\mid$$$$=\frac{\mid cos\theta \mid }{2}ln\mid \frac{tan\theta +\frac{1}{\mid cos\theta \mid }}{tan\theta -\frac{1}{\mid cos\theta \mid }}\mid =\frac{\mid cos\theta \mid }{2}ln\mid \frac{\xi \left(\theta \right)sin\theta +1}{\xi \left(\theta \right)-1}\mid with\xi \left(\theta \right)=\frac{cos\theta }{\mid cos\theta \mid }=\stackrel{-}{+}1$$

Commented by maxmathsup by imad last updated on 01/Oct/18

$$4)wehavef\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+2xt-1}\Rightarrow f^{{}^{\prime }}\left(x\right)=-{\int }_0^{\mathrm{\infty }}\frac{2t}{{(t^2+2xt-1)}^2}dt\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{t}{{(t^2+2xt-1)}^2}dt=-\frac{1}{2}{f}^{{}^{\prime }}\left(x\right)$$$$f\left(x\right)=\frac{1}{2}{(1+x^2)}^{-\frac{1}{2}}\{ln\mid x+\sqrt{1+x^2}\mid -ln\mid x-\sqrt{1+x^2}\mid \}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{-x}{2}{(1+x^2)}^{-\frac{3}{2}}\{ln\mid x+\sqrt{1+x^2\mid }-ln\mid x-\sqrt{1+x^2}\mid \}$$$$+\frac{1}{2\sqrt{1+x^2}}\{\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}-\frac{1-\frac{x}{\sqrt{1+x^2}}}{x-\sqrt{1+x^2}}\}$$$$=-\frac{x}{2(1+x^2)\sqrt{1+x^2}}\{ln\mid x+\sqrt{1+x^2}-ln\mid x-\sqrt{1+x^2}\mid \}$$$$+\frac{1}{2\sqrt{1+x^2}}\{\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\}\Rightarrow g\left(x\right)=\frac{x}{2(1+x^2)\sqrt{1+x^2}}ln\mid \frac{x+\sqrt{1+x^2}}{x-\sqrt{1+x^2}}\mid$$$$+\frac{1}{1+x^2}$$

Commented by maxmathsup by imad last updated on 01/Oct/18

$$5){\int }_0^{\mathrm{\infty }}\frac{tdt}{{(t^2+4t-1)}^2}=g\left(2\right)$$$$=\frac{1}{10\sqrt{5}}ln\mid \frac{2+\sqrt{5}}{2-\sqrt{5}}\mid -\frac{1}{10}.$$

Commented by maxmathsup by imad last updated on 01/Oct/18

$$g\left(x\right)=-\frac{1}{2}{f}^{{}^{\prime }}\left(x\right)\Rightarrow g\left(x\right)=\frac{x}{4(1+x^2)\sqrt{1+x^2}}ln\mid \frac{x+\sqrt{1+x^2}}{x-\sqrt{1+x^2}}\mid -\frac{1}{2(1+x^2)}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com