Compute the following indefinite integral:

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

Oh boy! Good mathematical Olympiad or any contest problems are hard to come by and this one particularly helps to cultivate analytical thinking and this is also one completely unpredictable and not straightforward problem.

This problem is by all means designed specifically for students to apply the substitution method.

The hardest part is to decide which part of the integrand should be replaced by the substitution. But we have to try, diligently and persistently as we know perseverance will paid off in the end.

Fortunately one thing that we do know is we can't simply let $u=\sqrt[4]{x^{10}+x^8+1}$, if we have that and when we compute the differential, we get:

$dx=\dfrac{2(x^{10}+x^8+1)^{\frac{3}{4}}du}{x^7(5x^2+4)}=\dfrac{2(u^3)du}{x^7(5x^2+4)}$

$\implies\,\displaystyle\int \dfrac{(3x^{10}+2x^8-2)\sqrt[4]{x^{10}+x^8+1}}{x^6} \,dx=\int \dfrac{(3x^{10}+2x^8-2)\sqrt[4]{u}}{x^6} \cdot \dfrac{2u^3du}{x^7(5x^2+4)}\,du$

This gives us a more headache integrand and it hence complicates the problem even more. Thus,

[MATH]\color{yellow}\bbox[5px,purple]{u=\sqrt[4]{x^{10}+x^8+1}}[/MATH] is a wrong substitution.

What we desperately need is for a substitution, says, $u$, that after we found its differential, we would get back something like $dx=\dfrac{\text{(some quantity)}du}{(3x^{10}+2x^8-2)}$ so that the first factor in the numerator of the given integrand can be canceled out.

Since we can't let [MATH]\color{yellow}\bbox[5px,purple]{u=\sqrt[4]{x^{10}+x^8+1}}[/MATH], let us try something like [MATH]\color{yellow}\bbox[5px,green]{u=\dfrac{\sqrt[4]{x^{10}+x^8+1}}{x}}[/MATH], then

$du=\dfrac{x\left(\dfrac{10x^9+8x^7}{4(x^{10}+x^8+1)^{\frac{3}{4}}}\right)-(x^{10}+x^8+1)^{\frac{1}{4}}}{x^2}\,dx$

$\,\,\,\,\,\,\,\,=\dfrac{10x^{10}+8x^8-4(x^{10}+x^8+1)}{4x^2(x^{10}+x^8+1)^{\frac{3}{4}}}\,dx$

$\,\,\,\,\,\,\,\,=\dfrac{6x^{10}+4x^8-4}{4x^2(x^{10}+x^8+1)^{\frac{3}{4}}}\,dx$

$\,\,\,\,\,\,\,\,=\dfrac{3x^{10}+2x^8-2}{2x^2(x^{10}+x^8+1)^{\frac{3}{4}}}\,dx$

$\,\,\,\,\,\,\,\,=\dfrac{3x^{10}+2x^8-2}{2x^2(\sqrt[4]{x^{10}+x^8+1})^3}\,dx$

$\,\,\,\,\,\,\,\,=\dfrac{3x^{10}+2x^8-2}{2x^2\left(\dfrac{\sqrt[4]{x^{10}+x^8+1}}{x}\right)^3x^3}\,dx$

$\,\,\,\,\,\,\,\,=\dfrac{3x^{10}+2x^8-2}{2x^5u^3}\,dx$

This brings $dx=\dfrac{2x^5u^3du}{3x^{10}+2x^8-2}$.

This differential looks promising and we will hence continue in this direction by replacing the pieces in the original integral:

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

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

[MATH]=\int \dfrac{(3x^{10}+2x^8-2)}{x^5}\cdot u \cdot \dfrac{2x^5u^3du}{(3x^{10}+2x^8-2)}[/MATH]

[MATH]=\int \dfrac{\cancel{(3x^{10}+2x^8-2)}}{\cancel{x^5}}\cdot u \cdot \dfrac{2\cancel{x^5}u^3du}{\cancel{(3x^{10}+2x^8-2)}}[/MATH]

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

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

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

[MATH]=\dfrac{2(x^{10}+x^8+1)^{\frac{5}{4}}}{5x^5}+C[/MATH]

If you are hard working and brilliant student, one way of tackling an intriguing problem could not satisfy your thirst for knowledge, yes, there is another way to attacking this great problem, and I will let you have some fun to explore it and I will be write back soon to share with you the other method. See you!

## No comments:

## Post a Comment