What is the integral of ((x^3-4x))/(3x) ?

The integral of ((x^3-4x))/(3x) is 1/3 ((x^3)/3-4x)+C

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integral of ((x^3-4x))/(3x)

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

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

\int \frac{4 x}{x} d x = 4 x

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

Add a constant to the solution

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

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What is the integral of ((x^3-4x))/(3x) ?
The integral of ((x^3-4x))/(3x) is 1/3 ((x^3)/3-4x)+C