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

The integral of (x^4)/(4x) is (x^4)/(16)+C

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

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

\frac{x ^{4}}{16} + C

Solution steps

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

Take the constant out:

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

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

Cancel the common factor:

x

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

Apply the Power Rule

: \frac{x ^{4}}{4}

= \frac{1}{4} \cdot \frac{x ^{4}}{4}

Simplify

\frac{1}{4} \cdot \frac{x ^{4}}{4} : \frac{x ^{4}}{16}

= \frac{x ^{4}}{16}

Add a constant to the solution

= \frac{x ^{4}}{16} + C

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