What is the integral of (\sqrt[5]{17}-(x^3)/7)^5x^2 ?

The integral of (\sqrt[5]{17}-(x^3)/7)^5x^2 is

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integral of (\sqrt[5]{17}-(x^3)/7)^5x^2

Solution steps

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Apply the Sum Rule:

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

\int 17 x^{2} d x = \frac{17 x ^{3}}{3}

\int \frac{5 \cdot 1 7 ^{\frac{4}{5}} x ^{5}}{7} d x = \frac{5 \cdot 1 7 ^{\frac{4}{5}} x ^{6}}{42}

\int \frac{10 \cdot 1 7 ^{\frac{3}{5}} x ^{8}}{49} d x = \frac{10 \cdot 1 7 ^{\frac{3}{5}} x ^{9}}{441}

\int \frac{10 \cdot 1 7 ^{\frac{2}{5}} x ^{11}}{343} d x = \frac{5 \cdot 1 7 ^{\frac{2}{5}} x ^{12}}{2058}

\int \frac{x ^{17}}{16807} d x = \frac{x ^{18}}{302526}

Add a constant to the solution

n = 3 \sum \infty \frac{1}{3} (\frac{2}{5})^{n}

\int \frac{1}{5 ^{4 x}} d x

\frac{d}{d x} (\frac{3 cos ( x )}{3 x ^{2} + 3 x - 1})

- 3 y^{^{'}} + \frac{2 y}{x} = \frac{( y ^{4} )}{x ^{4}}

\frac{d y}{d x} = \frac{x - y - 3}{x - 3}

What is the integral of (\sqrt[5]{17}-(x^3)/7)^5x^2 ?
The integral of (\sqrt[5]{17}-(x^3)/7)^5x^2 is