What is the integral of (6x^2)((x^1)/4) ?

The integral of (6x^2)((x^1)/4) is (3x^4)/8+C

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integral of (6x^2)((x^1)/4)

\int (6 x^{2}) (\frac{x ^{1}}{4}) d x

\frac{3 x ^{4}}{8} + C

Solution steps

\int 6 x^{2} \frac{x ^{1}}{4} d x

Take the constant out:

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

= 6 \cdot \frac{1}{4} \cdot \int x^{2} x^{1} d x

6 \cdot \frac{1}{4} = \frac{3}{2}

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

Simplify

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

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

Apply the Power Rule

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

= \frac{3}{2} \cdot \frac{x ^{4}}{4}

Simplify

\frac{3}{2} \cdot \frac{x ^{4}}{4} : \frac{3 x ^{4}}{8}

= \frac{3 x ^{4}}{8}

Add a constant to the solution

= \frac{3 x ^{4}}{8} + C

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What is the integral of (6x^2)((x^1)/4) ?
The integral of (6x^2)((x^1)/4) is (3x^4)/8+C