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 144187 by ArielVyny last updated on 22/Jun/21

∫_0 ^(+∞) (u^2 /(u^8 +2u^4 +1))du

$$\int_{\mathrm{0}} ^{+\infty} \frac{{u}^{\mathrm{2}} }{{u}^{\mathrm{8}} +\mathrm{2}{u}^{\mathrm{4}} +\mathrm{1}}{du} \\ $$

Answered by Ar Brandon last updated on 22/Jun/21

I=∫_0 ^∞ (u^2 /(u^8 +2u^4 +1))du=∫_0 ^∞ (u^2 /((u^4 +1)^2 ))du    =(1/4)∫_0 ^∞ (x^(−(1/4)) /((x+1)^2 ))dx=(1/4)β((3/4),(5/4))=(1/(16))β((3/4),(1/4))    =(1/(16))∙(π/(sin((π/4))))=(((√2)π)/(16))

$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{u}^{\mathrm{8}} +\mathrm{2u}^{\mathrm{4}} +\mathrm{1}}\mathrm{du}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{\mathrm{2}} }{\left(\mathrm{u}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{du} \\ $$ $$\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\beta\left(\frac{\mathrm{3}}{\mathrm{4}},\frac{\mathrm{5}}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{16}}\beta\left(\frac{\mathrm{3}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$ $$\:\:=\frac{\mathrm{1}}{\mathrm{16}}\centerdot\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)}=\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{16}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com