What is the integral of (-5x^4-4\sqrt[3]{x})/(4x) ?

The integral of (-5x^4-4\sqrt[3]{x})/(4x) is 1/4 (-(5x^4)/4-12\sqrt[3]{x})+C

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integral of (-5x^4-4\sqrt[3]{x})/(4x)

Solution steps

Take the constant out:

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

Expand

= \frac{1}{4} \cdot \int - 5 x^{3} - \frac{4}{x ^{\frac{2}{3}}} d x

Apply the Sum Rule:

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

= \frac{1}{4} (- \int 5 x^{3} d x - \int \frac{4}{x ^{\frac{2}{3}}} d x)

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

\int \frac{4}{x ^{\frac{2}{3}}} d x = 12 x^{\frac{1}{3}}

= \frac{1}{4} (- \frac{5 x ^{4}}{4} - 12 x^{\frac{1}{3}})

Apply radical rule:

Add a constant to the solution

\int 2 d x

x \to 3 + lim (\frac{- 5}{x - 3})

\frac{d}{d x} (\frac{e ^{x}}{4 - 3 e ^{x}})

\frac{\partial}{\partial x} (x^{2} y + x + 1)

e^{5 x}

What is the integral of (-5x^4-4\sqrt[3]{x})/(4x) ?
The integral of (-5x^4-4\sqrt[3]{x})/(4x) is 1/4 (-(5x^4)/4-12\sqrt[3]{x})+C