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

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

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

\frac{3}{7} x^{\frac{7}{3}} - 3 x^{\frac{4}{3}} + C

Solution steps

Expand

Apply the Sum Rule:

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

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

= \frac{3}{7} x^{\frac{7}{3}} - 3 x^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} x^{\frac{7}{3}} - 3 x^{\frac{4}{3}} + C

\int 4 x^{3} + 1 d x

f (x) = x^{3} - 9 x (3)

f (x) = tan^{2} (x), at x = \frac{- π}{4}

\frac{\partial}{\partial x} ((2 x - 5 y)^{2})

\int y^{3} + 10 x y^{4} - 2 x d x

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