$\displaystyle \int \left(x^{3m}+x^{2m}+x^m\right).\left(2x^{2m}+3x^{m}+6\right)^{\frac{1}{m}}dx$
Where $m\in\mathbb{N}$ and $x>0$
o primitiva(2)
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stuart clark
- Mesaje: 52
- Membru din: Mie Mar 02, 2011 5:20 pm
Re: o primitiva(2)
[quote="stuart clark"]$\displaystyle \int \left(x^{3m}+x^{2m}+x^m\right).\left(2x^{2m}+3x^{m}+6\right)^{\frac{1}{m}}dx$
$\displaystyle \int \left(x^{3m-1}+x^{2m-1}+x^{m-1}\right).\left(2x^{3m}+3x^{2m}+6x^{m}\right)^{\frac{1}{m}}dx$
Let $\displaystyle \left(2x^{3m}+3x^{2m}+6x^{m}\right) = t$
and $\displaystyle 6m\left(x^{3m-1}+x^{2m-1}+x^{m-1}\right)dx = dt$
$\displaystyle \frac{1}{6m}\int t^{\frac{1}{m}}dt = \frac{1}{6m}\frac{t^{\frac{1}{m}+1}}{\frac{1}{m}+1}+C$
$\displaystyle \int \left(x^{3m-1}+x^{2m-1}+x^{m-1}\right).\left(2x^{3m}+3x^{2m}+6x^{m}\right)^{\frac{1}{m}}dx$
Let $\displaystyle \left(2x^{3m}+3x^{2m}+6x^{m}\right) = t$
and $\displaystyle 6m\left(x^{3m-1}+x^{2m-1}+x^{m-1}\right)dx = dt$
$\displaystyle \frac{1}{6m}\int t^{\frac{1}{m}}dt = \frac{1}{6m}\frac{t^{\frac{1}{m}+1}}{\frac{1}{m}+1}+C$