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

The integral of (3x+4)^{1/2} is 2/9 (3x+4)^{3/2}+C

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

\int (3 x + 4)^{\frac{1}{2}} d x

\frac{2}{9} (3 x + 4)^{\frac{3}{2}} + C

Solution steps

\int (3 x + 4)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{3} d u

= \int \frac{u ^{\frac{1}{2}}}{3} d u

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x + 4

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

Simplify

\frac{1}{3} \cdot \frac{2}{3} (3 x + 4)^{\frac{3}{2}} : \frac{2}{9} (3 x + 4)^{\frac{3}{2}}

= \frac{2}{9} (3 x + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (3 x + 4)^{\frac{3}{2}} + C

(x y^{2} + 3 y^{2}) d y - 2 x d x = 0

\frac{d}{d x} (\frac{5}{4 + x})

(4, 1), (7, 10)

\frac{d}{d x} (\frac{18}{x})

\frac{\partial}{\partial y} (2 cos (x) + cos (x + y))

What is the integral of (3x+4)^{1/2} ?
The integral of (3x+4)^{1/2} is 2/9 (3x+4)^{3/2}+C