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

The integral of (6x^2)/(sqrt(x)) is 12/5 x^{5/2}+C

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integral of (6x^2)/(sqrt(x))

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

\frac{12}{5} x^{\frac{5}{2}} + C

Solution steps

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

Take the constant out:

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

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

Simplify

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

= 6 \cdot \int x^{\frac{3}{2}} d x

Apply the Power Rule

= 6 \cdot \frac{2}{5} x^{\frac{5}{2}}

Simplify

6 \cdot \frac{2}{5} x^{\frac{5}{2}} : \frac{12}{5} x^{\frac{5}{2}}

= \frac{12}{5} x^{\frac{5}{2}}

Add a constant to the solution

= \frac{12}{5} x^{\frac{5}{2}} + C

What is the integral of (6x^2)/(sqrt(x)) ?
The integral of (6x^2)/(sqrt(x)) is 12/5 x^{5/2}+C