What is the integral of (4-x^{2/3})^3 ?

The integral of (4-x^{2/3})^3 is 64x-144/5 x^{5/3}+36/7 x^{7/3}-(x^3)/3+C

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integral of (4-x^{2/3})^3

\int (4 - x^{\frac{2}{3}})^{3} d x

64 x - \frac{144}{5} x^{\frac{5}{3}} + \frac{36}{7} x^{\frac{7}{3}} - \frac{x ^{3}}{3} + C

Solution steps

\int (4 - x^{\frac{2}{3}})^{3} d x

Expand

(4 - x^{\frac{2}{3}})^{3} : 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2}

= \int 64 - 48 x^{\frac{2}{3}} + 12 x^{\frac{4}{3}} - x^{2} 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 64 d x - \int 48 x^{\frac{2}{3}} d x + \int 12 x^{\frac{4}{3}} d x - \int x^{2} d x

\int 64 d x = 64 x

\int 48 x^{\frac{2}{3}} d x = \frac{144}{5} x^{\frac{5}{3}}

\int 12 x^{\frac{4}{3}} d x = \frac{36}{7} x^{\frac{7}{3}}

\int x^{2} d x = \frac{x ^{3}}{3}

= 64 x - \frac{144}{5} x^{\frac{5}{3}} + \frac{36}{7} x^{\frac{7}{3}} - \frac{x ^{3}}{3}

Add a constant to the solution

= 64 x - \frac{144}{5} x^{\frac{5}{3}} + \frac{36}{7} x^{\frac{7}{3}} - \frac{x ^{3}}{3} + C

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What is the integral of (4-x^{2/3})^3 ?
The integral of (4-x^{2/3})^3 is 64x-144/5 x^{5/3}+36/7 x^{7/3}-(x^3)/3+C