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 158591 by cortano last updated on 06/Nov/21

I=∫ (dx/(1+x^6 )) =?

$$\:{I}=\int\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{6}} }\:=? \\ $$

Commented by tounghoungko last updated on 06/Nov/21

I=∫ ((x^2 +1)/(x^6 +1)) dx−∫ (x^2 /(x^6 +1)) dx I=∫ ((x^2 +1)/((x^2 +1)(x^4 −x^2 +1))) dx−∫ (x^2 /((x^3 )^2 +1)) dx I=(1/2)∫ ((x^2 +1)/(x^2 ((x−(1/x))^2 +1)))dx −(1/2)∫ ((x^2 −1)/(x^2 ((x+(1/x))^2 −3)))dx −(1/3)∫ ((3x^2 )/((x^3 )^2 +1))dx

$${I}=\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}\:{dx}−\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{6}} +\mathrm{1}}\:{dx} \\ $$$${I}=\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)}\:{dx}−\int\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} \left(\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{1}\right)}{dx}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \left(\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{3}\right)}{dx}\:−\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{\mathrm{3}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$ \\ $$

Commented by ilhamdiii last updated on 06/Nov/21

oke

$${oke} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com