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Popular >

integral of (y-1)/y

Solution

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

Solution

y - ln ∣ y ∣ + C

Solution steps

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

Expand

\frac{y - 1}{y} : 1 - \frac{1}{y}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d y = y

\int \frac{1}{y} d y = ln ∣ y ∣

= y - ln ∣ y ∣

Add a constant to the solution

= y - ln ∣ y ∣ + C

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Frequently Asked Questions (FAQ)

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