Question Number 133963 by bobhans last updated on 25/Feb/21
$$\mathcal{Y}\:=\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{6}}]{\mathrm{1}+{x}^{\mathrm{6}} }}? \\ $$
Answered by EDWIN88 last updated on 26/Feb/21
![Y= ∫ (dx/(x ((1+x^(−6) ))^(1/6) )) = ∫ (x^6 /(x^7 ((1+x^(−6) ))^(1/6) )) dx let u^6 = 1+x^(−6) ⇒ u^5 du = −x^(−7) dx ⇒ (dx/x^7 ) = −u^5 du and x^6 = (1/(u^6 −1)) Y = −∫ (1/(u^6 −1)) . ((u^5 du )/u) =−∫ ((u^4 du)/(u^6 −1)) Use Partial fraction Y =(1/6)[∫ ((u−1)/(u^2 +u+1)) du−∫ ((u+1)/(u^2 −u+1)) du +∫ (du/(u+1)) −∫ (du/(u−1)) ] Y= (1/6)[ (1/2)ln (u^2 +u+1)−(√3) tan^(−1) (((2u+1)/( (√3))))− (1/2)ln ∣u^2 −u+1∣−(√3) tan^(−1) (((2u−1)/( (√3))))+ln ∣((u+1)/(u−1))∣ ]+ c](https://www.tinkutara.com/question/Q133967.png)
$$\mathbb{Y}=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}\:\sqrt[{\mathrm{6}}]{\mathrm{1}+\mathrm{x}^{−\mathrm{6}} }}\:=\:\int\:\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{x}^{\mathrm{7}} \:\sqrt[{\mathrm{6}}]{\mathrm{1}+\mathrm{x}^{−\mathrm{6}} }}\:\mathrm{dx}\: \\ $$$$\mathrm{let}\:\mathrm{u}^{\mathrm{6}} \:=\:\mathrm{1}+\mathrm{x}^{−\mathrm{6}} \:\Rightarrow\:\mathrm{u}^{\mathrm{5}} \:\mathrm{du}\:=\:−\mathrm{x}^{−\mathrm{7}} \:\mathrm{dx} \\ $$$$\:\Rightarrow\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{7}} }\:=\:−\mathrm{u}^{\mathrm{5}} \:\mathrm{du}\:\mathrm{and}\:\mathrm{x}^{\mathrm{6}} \:=\:\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{6}} −\mathrm{1}} \\ $$$$\mathbb{Y}\:=\:−\int\:\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{6}} −\mathrm{1}}\:.\:\frac{\mathrm{u}^{\mathrm{5}} \:\mathrm{du}\:}{\mathrm{u}}\:=−\int\:\frac{\mathrm{u}^{\mathrm{4}} \:\mathrm{du}}{\mathrm{u}^{\mathrm{6}} −\mathrm{1}} \\ $$$$\mathrm{Use}\:\mathrm{Partial}\:\mathrm{fraction} \\ $$$$\mathbb{Y}\:=\frac{\mathrm{1}}{\mathrm{6}}\left[\int\:\frac{\mathrm{u}−\mathrm{1}}{\mathrm{u}^{\mathrm{2}} +\mathrm{u}+\mathrm{1}}\:\mathrm{du}−\int\:\frac{\mathrm{u}+\mathrm{1}}{\mathrm{u}^{\mathrm{2}} −\mathrm{u}+\mathrm{1}}\:\mathrm{du}\:+\int\:\frac{\mathrm{du}}{\mathrm{u}+\mathrm{1}}\:−\int\:\frac{\mathrm{du}}{\mathrm{u}−\mathrm{1}}\:\right] \\ $$$$\mathbb{Y}=\:\frac{\mathrm{1}}{\mathrm{6}}\left[\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{u}^{\mathrm{2}} +\mathrm{u}+\mathrm{1}\right)−\sqrt{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2u}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\right. \\ $$$$\left.\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{u}^{\mathrm{2}} −\mathrm{u}+\mathrm{1}\mid−\sqrt{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2u}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\mathrm{ln}\:\mid\frac{\mathrm{u}+\mathrm{1}}{\mathrm{u}−\mathrm{1}}\mid\:\right]+\:\mathrm{c} \\ $$$$ \\ $$