What is the integral of ((x^3)/3)(1/x) ?

The integral of ((x^3)/3)(1/x) is (x^3)/9+C

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integral of ((x^3)/3)(1/x)

\int (\frac{x ^{3}}{3}) (\frac{1}{x}) d x

\frac{x ^{3}}{9} + C

Solution steps

\int \frac{x ^{3}}{3} \cdot \frac{1}{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 x^{3} \frac{1}{x} d x

Simplify

x^{3} \frac{1}{x} : x^{2}

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

Apply the Power Rule

: \frac{x ^{3}}{3}

= \frac{1}{3} \cdot \frac{x ^{3}}{3}

Simplify

\frac{1}{3} \cdot \frac{x ^{3}}{3} : \frac{x ^{3}}{9}

= \frac{x ^{3}}{9}

Add a constant to the solution

= \frac{x ^{3}}{9} + C

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What is the integral of ((x^3)/3)(1/x) ?
The integral of ((x^3)/3)(1/x) is (x^3)/9+C