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

The integral of \sqrt[3]{x}-4x is 3/4 x^{4/3}-2x^2+C

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

\frac{3}{4} x^{\frac{4}{3}} - 2 x^{2} + C

Solution steps

Apply the Sum Rule:

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

\int 4 x d x = 2 x^{2}

= \frac{3}{4} x^{\frac{4}{3}} - 2 x^{2}

Add a constant to the solution

= \frac{3}{4} x^{\frac{4}{3}} - 2 x^{2} + C

x \to 0 lim ((1 + a x)^{\frac{1}{x}})

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

(6 x + e^{- x^{2}})^{4}

(\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y - 7 y^{3})

\int_{a}^{b} 25 - x^{2} d x

What is the integral of \sqrt[3]{x}-4x ?
The integral of \sqrt[3]{x}-4x is 3/4 x^{4/3}-2x^2+C