What is the integral of 1/(-2y) ?

The integral of 1/(-2y) is -1/2 ln|y|+C

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integral of 1/(-2y)

\int \frac{1}{- 2 y} d y

- \frac{1}{2} ln ∣ y ∣ + C

Solution steps

\int \frac{1}{- 2 y} d y

Take the constant out:

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

= \frac{1}{- 2} \cdot \int \frac{1}{y} d y

Simplify

= (- \frac{1}{2}) \cdot \int \frac{1}{y} d y

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= (- \frac{1}{2}) ln ∣ y ∣

Simplify

= - \frac{1}{2} ln ∣ y ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ y ∣ + C

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