Hard and Intriguing Indefinite Integral Problem (Second Method)

On this blog post, I will show another intelligent way to tackle the problem(Hard and Intriguing Indefinite Integral Problem ) that to compute the following indefinite integral:

[MATH]\int \dfrac{(3x^{10}+2x^8-2)\sqrt[4]{x^{10}+x^8+1}}{x^6} \,dx[/MATH]

If we rewrite the integrand such that we have:

[MATH]\dfrac{(3x^{10}+2x^8-2)\sqrt[4]{x^{10}+x^8+1}}{x^6}[/MATH]

[MATH]=\dfrac{(3x^{10}+2x^8-2)}{x^5}\cdot\dfrac{\sqrt[4]{x^{10}+x^8+1}}{x}[/MATH]

[MATH]=\left(\dfrac{3x^{10}}{x^5}+\dfrac{2x^{8}}{x^5}-\dfrac{2}{x^5}\right)\cdot\sqrt[4]{\dfrac{x^{10}+x^8+1}{x^4}}[/MATH]

[MATH]=\left(3x^5+2x^3-2x^{-5}\right)\cdot\sqrt[4]{x^{6}+x^4+x^{-4}}[/MATH]

We see that we could use the substitution method and let $u^4=x^{6}+x^4+x^{-4}$ so that

$4u^3\dfrac{du}{dx}=6x^{5}+4x^2-4x^{-5}$

$\dfrac{2u^3du}{3x^{5}+2x^2-2x^{-5}}=dx$

The integral is then

[MATH]\int \dfrac{(3x^{10}+2x^8-2)\sqrt[4]{x^{10}+x^8+1}}{x^6} \,dx[/MATH]

[MATH]=\int \left(3x^5+2x^3-2x^{-5}\right)\cdot\sqrt[4]{x^{6}+x^4+x^{-4}} \,dx[/MATH]

[MATH]=\int \cancel{(3x^5+2x^3-2x^{-5})}\cdot\sqrt[4]{u^4} \,\dfrac{2u^3du}{\cancel{(3x^5+2x^3-2x^{-5})}}[/MATH]

[MATH]=\int 2u^4 \,du[/MATH]

[MATH]=\dfrac{2u^5}{5}+C [/MATH]

[MATH]=\dfrac{2(\sqrt[4]{x^{6}+x^4+x^{-4}})^5}{5}+C [/MATH]

[MATH]=\dfrac{2(x^6+x^4+x^{-4})^{\frac{5}{4}}}{5}+C [/MATH]

Kudos to you if you have solved this problem and I would be very happy if you could share with me your method and together, we can learn new strategies that will make us even stronger!