What is the integral of \sqrt[3]{(8-x)^4} ?

The integral of \sqrt[3]{(8-x)^4} is -3/7 (8-x)^{7/3}+C

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integral of \sqrt[3]{(8-x)^4}

- \frac{3}{7} (8 - x)^{\frac{7}{3}} + C

Solution steps

Apply u-substitution

: \int - u^{\frac{4}{3}} d u

= \int - u^{\frac{4}{3}} d u

Take the constant out:

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

= - \int u^{\frac{4}{3}} d u

Apply the Power Rule

: \frac{3}{7} u^{\frac{7}{3}}

= - \frac{3}{7} u^{\frac{7}{3}}

Substitute back

u = 8 - x

= - \frac{3}{7} (8 - x)^{\frac{7}{3}}

Add a constant to the solution

= - \frac{3}{7} (8 - x)^{\frac{7}{3}} + C

\int 3 x + 2 d x

\frac{( x ^{2} + 3 ) d y}{d x} = x y, y (1) = 8

4 ln (x)

\frac{\partial}{\partial u} (u v - w)

2 x^{2} + 3 x, at x = 1

What is the integral of \sqrt[3]{(8-x)^4} ?
The integral of \sqrt[3]{(8-x)^4} is -3/7 (8-x)^{7/3}+C