What is the integral of 8x^{1/3}+6x^{-1/3} ?

The integral of 8x^{1/3}+6x^{-1/3} is 6x^{4/3}+9x^{2/3}+C

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integral of 8x^{1/3}+6x^{-1/3}

\int 8 x^{\frac{1}{3}} + 6 x^{- \frac{1}{3}} d x

6 x^{\frac{4}{3}} + 9 x^{\frac{2}{3}} + C

Solution steps

\int 8 x^{\frac{1}{3}} + 6 x^{- \frac{1}{3}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 8 x^{\frac{1}{3}} d x + \int 6 x^{- \frac{1}{3}} d x

\int 8 x^{\frac{1}{3}} d x = 6 x^{\frac{4}{3}}

\int 6 x^{- \frac{1}{3}} d x = 9 x^{\frac{2}{3}}

= 6 x^{\frac{4}{3}} + 9 x^{\frac{2}{3}}

Add a constant to the solution

= 6 x^{\frac{4}{3}} + 9 x^{\frac{2}{3}} + C

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What is the integral of 8x^{1/3}+6x^{-1/3} ?
The integral of 8x^{1/3}+6x^{-1/3} is 6x^{4/3}+9x^{2/3}+C