What is the integral of (7^{2/3}-x^{2/3})^3 ?

The integral of (7^{2/3}-x^{2/3})^3 is 49x-(63*7^{1/3})/5 x^{5/3}+9/(7^{1/3)}x^{7/3}-(x^3)/3+C

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integral of (7^{2/3}-x^{2/3})^3

\int (7^{\frac{2}{3}} - x^{\frac{2}{3}})^{3} d x

49 x - \frac{63 \cdot 7 ^{\frac{1}{3}}}{5} x^{\frac{5}{3}} + \frac{9}{7 ^{\frac{1}{3}}} x^{\frac{7}{3}} - \frac{x ^{3}}{3} + C

Solution steps

\int (7^{\frac{2}{3}} - x^{\frac{2}{3}})^{3} d x

Expand

(7^{\frac{2}{3}} - x^{\frac{2}{3}})^{3} : 49 - 3 \cdot 7^{\frac{4}{3}} x^{\frac{2}{3}} + 3 \cdot 7^{\frac{2}{3}} x^{\frac{4}{3}} - x^{2}

= \int 49 - 3 \cdot 7^{\frac{4}{3}} x^{\frac{2}{3}} + 3 \cdot 7^{\frac{2}{3}} 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 49 d x - \int 3 \cdot 7^{\frac{4}{3}} x^{\frac{2}{3}} d x + \int 3 \cdot 7^{\frac{2}{3}} x^{\frac{4}{3}} d x - \int x^{2} d x

\int 49 d x = 49 x

\int 3 \cdot 7^{\frac{4}{3}} x^{\frac{2}{3}} d x = \frac{63 \cdot 7 ^{\frac{1}{3}}}{5} x^{\frac{5}{3}}

\int 3 \cdot 7^{\frac{2}{3}} x^{\frac{4}{3}} d x = \frac{9}{7 ^{\frac{1}{3}}} x^{\frac{7}{3}}

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

= 49 x - \frac{63 \cdot 7 ^{\frac{1}{3}}}{5} x^{\frac{5}{3}} + \frac{9}{7 ^{\frac{1}{3}}} x^{\frac{7}{3}} - \frac{x ^{3}}{3}

Add a constant to the solution

= 49 x - \frac{63 \cdot 7 ^{\frac{1}{3}}}{5} x^{\frac{5}{3}} + \frac{9}{7 ^{\frac{1}{3}}} x^{\frac{7}{3}} - \frac{x ^{3}}{3} + C

What is the integral of (7^{2/3}-x^{2/3})^3 ?
The integral of (7^{2/3}-x^{2/3})^3 is 49x-(63*7^{1/3})/5 x^{5/3}+9/(7^{1/3)}x^{7/3}-(x^3)/3+C