What is the integral of 1/(sqrt(3x-2)) ?

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

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integral of 1/(sqrt(3x-2))

\int \frac{1}{3 x - 2} d x

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

Solution steps

\int \frac{1}{3 x - 2} 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 - 2

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

Simplify

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

= \frac{2}{3} 3 x - 2

Add a constant to the solution

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

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

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