What is the integral of 8x^{1/3}-4x^{-1/3} ?

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

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integral of 8x^{1/3}-4x^{-1/3}

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

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

Solution steps

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

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

\int 4 x^{- \frac{1}{3}} d x = 6 x^{\frac{2}{3}}

= 6 x^{\frac{4}{3}} - 6 x^{\frac{2}{3}}

Add a constant to the solution

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

x \to - \infty lim (\frac{2 e ^{- \frac{1}{2} x}}{x ^{3}})

\int 2 cos^{5} (x) sin^{4} (x) d x

f (x) = - \frac{9 x ^{2}}{2} + 8 x - 22

\frac{d h}{d t} = - h^{\frac{1}{3}}

\int (x^{2})^{23} d (x^{2})^{2}

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