Menu Close

x-4-1-x-6-1-dx-




Question Number 84733 by manr last updated on 15/Mar/20
∫((x^4 +1)/(x^6 +1))dx
$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx} \\ $$
Answered by TANMAY PANACEA last updated on 15/Mar/20
  ∫(((x^2 +1)^2 −2x^2 )/((x^2 +1)(x^4 −x^2 +1)))dx  ∫((x^2 +1)/(x^4 −x^2 +1))−2∫((x^2 dx)/(x^6 +1))  ∫((1+(1/x^2 ))/((x^2 +(1/x^2 )−1)))−(2/3)∫((d(x^3 ))/((x^3 )^2 +1))  I=I_1 −I_2   I_1 =∫((d(x−(1/x)))/((x−(1/x))^2 +1))=tan^(−1) (x−(1/x))+C_1   I_2 =(2/3)∫((d(x^3 ))/((x^3 )^2 +1))=(2/3)tan^(−1) (x^3 )  I=tan^− (x−(1/x))+(2/3)tan^(−1) (x^3 )+C
$$ \\ $$$$\int\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}\int\frac{{x}^{\mathrm{2}} {dx}}{{x}^{\mathrm{6}} +\mathrm{1}} \\ $$$$\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{d}\left({x}^{\mathrm{3}} \right)}{\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${I}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{1}}={tan}^{−\mathrm{1}} \left({x}−\frac{\mathrm{1}}{{x}}\right)+{C}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{d}\left({x}^{\mathrm{3}} \right)}{\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{3}}{tan}^{−\mathrm{1}} \left({x}^{\mathrm{3}} \right) \\ $$$${I}={tan}^{−} \left({x}−\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{2}}{\mathrm{3}}{tan}^{−\mathrm{1}} \left({x}^{\mathrm{3}} \right)+{C} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *