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Find-x-7-x-5-x-3-x-1-x-10-dx-x-R-




Question Number 156808 by MathSh last updated on 15/Oct/21
Find:  𝛀 =∫ ((x^7  - x^5  + x^3  - x)/(1 + x^(10) )) dx  ;  x∈R
$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\int\:\frac{\mathrm{x}^{\mathrm{7}} \:-\:\mathrm{x}^{\mathrm{5}} \:+\:\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{x}}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{10}} }\:\mathrm{dx}\:\:;\:\:\mathrm{x}\in\mathbb{R} \\ $$
Commented by tabata last updated on 15/Oct/21
𝛀 = ∫ ((x^7 −x^5 +x^3 −x)/(1+x^(10) ))dx    𝛀 = ∫ (−(2/5)) ((2x)/((x^2 +1))) + ((4x^7 −3x^5 +2x^3 −x)/(5(x^8 −x^6 +x^4 −x^2 +1))) dx    𝛀 = − (2/5) ln ∣ x^2 +1∣ + (1/(10)) ln ∣ x^8 −x^6 +x^4 −x^2 +1∣ + C    ⟨ M . T  ⟩
$$\boldsymbol{\Omega}\:=\:\int\:\frac{\boldsymbol{{x}}^{\mathrm{7}} −\boldsymbol{{x}}^{\mathrm{5}} +\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{x}}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{10}} }\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{\Omega}\:=\:\int\:\left(−\frac{\mathrm{2}}{\mathrm{5}}\right)\:\frac{\mathrm{2}\boldsymbol{{x}}}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)}\:+\:\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{7}} −\mathrm{3}\boldsymbol{{x}}^{\mathrm{5}} +\mathrm{2}\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{x}}}{\mathrm{5}\left(\boldsymbol{{x}}^{\mathrm{8}} −\boldsymbol{{x}}^{\mathrm{6}} +\boldsymbol{{x}}^{\mathrm{4}} −\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right)}\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{\Omega}\:=\:−\:\frac{\mathrm{2}}{\mathrm{5}}\:\boldsymbol{{ln}}\:\mid\:\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\mid\:+\:\frac{\mathrm{1}}{\mathrm{10}}\:\boldsymbol{{ln}}\:\mid\:\boldsymbol{{x}}^{\mathrm{8}} −\boldsymbol{{x}}^{\mathrm{6}} +\boldsymbol{{x}}^{\mathrm{4}} −\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\mid\:+\:\boldsymbol{{C}} \\ $$$$ \\ $$$$\langle\:\boldsymbol{{M}}\:.\:\boldsymbol{{T}}\:\:\rangle \\ $$

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