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

The integral of ((5x^3-3\sqrt[5]{x})/(5x)) is 1/3 (x^3-9\sqrt[5]{x})+C

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

Solution steps

Take the constant out:

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

Expand

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

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

\int \frac{3}{x ^{\frac{4}{5}}} d x = 15 x^{\frac{1}{5}}

= \frac{1}{5} (\frac{5 x ^{3}}{3} - 15 x^{\frac{1}{5}})

Simplify

Add a constant to the solution

x \to 3 lim (\frac{x - 3}{x ^{2} - 9})

f (x) = (\frac{x + 7}{x + 9})^{6}

\int \frac{1}{4 x 16 x ^{2} - 25} d x

- 3 e^{- 3 x} + e^{4 x} \cdot 4

\int y^{2} e^{3 y} d y

What is the integral of ((5x^3-3\sqrt[5]{x})/(5x)) ?
The integral of ((5x^3-3\sqrt[5]{x})/(5x)) is 1/3 (x^3-9\sqrt[5]{x})+C