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))

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

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

Solution steps

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

Take the constant out:

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

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

Expand

\frac{- 5 x ^{4} - 4 3 x}{x} : - 5 x^{3} - \frac{4}{x ^{\frac{2}{3}}}

= \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:

a^{\frac{1}{n}} = n a

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

Add a constant to the solution

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

\frac{d y}{d x} + \frac{2}{x} y = 4 x

\int_{3}^{8} \frac{1}{x ^{2} - 1} d x

t an g e n t y = x^{2} - 6

x \to \infty lim (0 + x)

\frac{d}{d c} (\frac{sin ( 2 c )}{2 - cos ( 2 c )})

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