What is the integral of 1/(sqrt(3x+8)) ?

The integral of 1/(sqrt(3x+8)) is 2/3 sqrt(3x+8)+C

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integral of 1/(sqrt(3x+8))

\int \frac{1}{3 x + 8} d x

\frac{2}{3} 3 x + 8 + C

Solution steps

\int \frac{1}{3 x + 8} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = 3 x + 8

= \frac{1}{3} \cdot 2 (3 x + 8)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot 2 (3 x + 8)^{\frac{1}{2}} : \frac{2}{3} 3 x + 8

= \frac{2}{3} 3 x + 8

Add a constant to the solution

= \frac{2}{3} 3 x + 8 + C

\int x^{2} + 16 d x

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

5 x^{2} - 2 x

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{8} e^{- x} d y = 0

\frac{d}{d x} (- a^{2} sin (a x))

What is the integral of 1/(sqrt(3x+8)) ?
The integral of 1/(sqrt(3x+8)) is 2/3 sqrt(3x+8)+C