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

The integral of 1/((3x-1)) is 1/3 ln|3x-1|+C

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

integral

\frac{1}{( 3 x - 1 )}

\frac{1}{3} ln ∣ 3 x - 1 ∣ + C

Solution steps

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x - 1

= \frac{1}{3} ln ∣ 3 x - 1 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 x - 1 ∣ + C

x^{2} (x^{3} + 2)^{4}

\frac{( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )}

\frac{- cos ^{2} ( x )}{2}

1. 3^{x}

\frac{3 x}{( x ^{2} + 4 x + 6 )}

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