What is the integral of 1/(12-4y) ?

The integral of 1/(12-4y) is -1/4 ln|12-4y|+C

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integral of 1/(12-4y)

\int \frac{1}{12 - 4 y} d y

- \frac{1}{4} ln ∣ 12 - 4 y ∣ + C

Solution steps

\int \frac{1}{12 - 4 y} d y

Apply u-substitution

= \int - \frac{1}{4 u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = 12 - 4 y

= - \frac{1}{4} ln ∣ 12 - 4 y ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ 12 - 4 y ∣ + C

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What is the integral of 1/(12-4y) ?
The integral of 1/(12-4y) is -1/4 ln|12-4y|+C