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

The integral of (6x^2)(x^{1/4}) is 24/13 x^{13/4}+C

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

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

\frac{24}{13} x^{\frac{13}{4}} + C

Solution steps

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

Take the constant out:

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

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

Simplify

x^{2} x^{\frac{1}{4}} : x^{\frac{9}{4}}

= 6 \cdot \int x^{\frac{9}{4}} d x

Apply the Power Rule

: \frac{4}{13} x^{\frac{13}{4}}

= 6 \cdot \frac{4}{13} x^{\frac{13}{4}}

Simplify

6 \cdot \frac{4}{13} x^{\frac{13}{4}} : \frac{24}{13} x^{\frac{13}{4}}

= \frac{24}{13} x^{\frac{13}{4}}

Add a constant to the solution

= \frac{24}{13} x^{\frac{13}{4}} + C

\int \frac{1}{6 x + 2} d x

\int 6 x e^{x} d x

\int x^{\frac{3}{8}} d x

\int 1^{2} (2 x - 1)^{3} d x

What is the integral of (6x^2)(x^{1/4}) ?
The integral of (6x^2)(x^{1/4}) is 24/13 x^{13/4}+C