What is the integral of sqrt(3x+2) ?

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

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integral of sqrt(3x+2)

\int 3 x + 2 d x

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

Solution steps

\int 3 x + 2 d x

Apply u-substitution

: \int \frac{u}{3} d u

= \int \frac{u}{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 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 + 2

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

Simplify

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

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

Add a constant to the solution

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

\frac{( x ^{2} + 3 ) d y}{d x} = x y, y (1) = 8

4 ln (x)

\frac{\partial}{\partial u} (u v - w)

2 x^{2} + 3 x, at x = 1

f (x) = 5 x - 3

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