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)

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

\frac{1}{3} (x^{3} - 9 5 x) + C

Solution steps

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

Take the constant out:

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

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

Expand

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

= \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

\frac{1}{5} (\frac{5 x ^{3}}{3} - 15 x^{\frac{1}{5}}) : \frac{1}{3} (x^{3} - 9 5 x)

= \frac{1}{3} (x^{3} - 9 5 x)

Add a constant to the solution

= \frac{1}{3} (x^{3} - 9 5 x) + C

g (x) = 2 - x^{3}

\int \frac{x ^{3} + 8}{x + 2} d x

\int - t^{\frac{3}{2}} d t

10 e^{x}

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

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