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

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

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integral of (sqrt(x))/4

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

\frac{1}{6} x^{\frac{3}{2}} + C

Solution steps

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

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{2}{3} x^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} x^{\frac{3}{2}} : \frac{1}{6} x^{\frac{3}{2}}

= \frac{1}{6} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} x^{\frac{3}{2}} + C

\frac{\partial}{\partial y} (\frac{x}{y})

x \to 2 lim (\frac{( x - 2 )}{x})

(((- 3)^{^{'}})^{4})^{^{'}}

\frac{d}{d x} (\frac{x ^{4} + 7 x ^{3} + 14 x ^{2}}{x + 3})

\int 3 y^{2} d y

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