integrate 1/(x+x^(1/3))
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Originally Posted by twilightstr integrate 1/(x+x^(1/3)) put $\displaystyle x^{\frac{1}{3}} =t$ so $\displaystyle \frac{x^{\frac{-2}{3}}dx}{3} = dt$ thus $\displaystyle dx = 3t^2dt$ So $\displaystyle \int{\frac{dx}{x+x^{1/3}} } $ $\displaystyle = \int{\frac{3t^2dt}{t^3+t} } $ Go ahead!
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