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

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

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

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

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

Solution steps

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

Cancel the common factor:

x

= \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

x \to 2 lim (\frac{f ( x )}{g ( x )})

\int \frac{( x ^{2} - x + 6 )}{( x ^{3} - 3 x )} d x

x^{2} - 6

(2 + x) (\frac{d y}{d x}) = 3 y

f (x) = \frac{3 x}{x}, x = 8