What is the integral of (6x+2)^{-1} ?

The integral of (6x+2)^{-1} is 1/6 ln|6x+2|+C

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integral of (6x+2)^{-1}

\int (6 x + 2)^{- 1} d x

\frac{1}{6} ln ∣ 6 x + 2 ∣ + C

Solution steps

\int (6 x + 2)^{- 1} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \int \frac{1}{6 x + 2} d x

Apply u-substitution

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 6 x + 2

= \frac{1}{6} ln ∣ 6 x + 2 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ 6 x + 2 ∣ + C

What is the integral of (6x+2)^{-1} ?
The integral of (6x+2)^{-1} is 1/6 ln|6x+2|+C