What is the integral of 1/(1-6y) ?

The integral of 1/(1-6y) is -1/6 ln|1-6y|+C

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integral of 1/(1-6y)

\int \frac{1}{1 - 6 y} d y

- \frac{1}{6} ln ∣ 1 - 6 y ∣ + C

Solution steps

\int \frac{1}{1 - 6 y} d y

Apply u-substitution

: \int - \frac{1}{6 u} d u

= \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 = 1 - 6 y

= - \frac{1}{6} ln ∣ 1 - 6 y ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ 1 - 6 y ∣ + C

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