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

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

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

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

\frac{2}{3} (x + 4)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int u^{\frac{1}{2}} d u

= \int u^{\frac{1}{2}} d u

Apply the Power Rule

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

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x + 4

= \frac{2}{3} (x + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x + 4)^{\frac{3}{2}} + C

\frac{d y}{d x} = (2 x)

x \to \infty lim (\frac{4 x ^{2}}{e ^{4 x}})

\frac{\partial}{\partial x} (x^{3} + x^{2} \cdot x^{3} - 2 y^{2})

\frac{d}{d x} (\frac{x - 2}{x ^{2} + 1})

\frac{d}{d x} (- 16)

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