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

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

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

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

- \frac{1}{5} (7 - x^{3})^{\frac{5}{3}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{u ^{\frac{2}{3}}}{3} d u

= \int - \frac{u ^{\frac{2}{3}}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{3} \cdot \int u^{\frac{2}{3}} d u

Apply the Power Rule

: \frac{3}{5} u^{\frac{5}{3}}

= - \frac{1}{3} \cdot \frac{3}{5} u^{\frac{5}{3}}

Substitute back

u = 7 - x^{3}

= - \frac{1}{3} \cdot \frac{3}{5} (7 - x^{3})^{\frac{5}{3}}

Simplify

- \frac{1}{3} \cdot \frac{3}{5} (7 - x^{3})^{\frac{5}{3}} : - \frac{1}{5} (7 - x^{3})^{\frac{5}{3}}

= - \frac{1}{5} (7 - x^{3})^{\frac{5}{3}}

Add a constant to the solution

= - \frac{1}{5} (7 - x^{3})^{\frac{5}{3}} + C

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

x \to - \infty lim (x^{3} - 5 x^{2})

\int x^{2} \cdot x - 5 d x

y = x^{2} (x^{3} - 1)^{5}

y = 2 x + 1, at x = 4

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