What is the integral of (y-3)/(4y) ?

The integral of (y-3)/(4y) is 1/4 (y-3ln|y|)+C

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integral of (y-3)/(4y)

\int \frac{y - 3}{4 y} d y

\frac{1}{4} (y - 3 ln ∣ y ∣) + C

Solution steps

\int \frac{y - 3}{4 y} d y

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{y - 3}{y} d y

Expand

\frac{y - 3}{y} : 1 - \frac{3}{y}

= \frac{1}{4} \cdot \int 1 - \frac{3}{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

= \frac{1}{4} (\int 1 d y - \int \frac{3}{y} d y)

\int 1 d y = y

\int \frac{3}{y} d y = 3 ln ∣ y ∣

= \frac{1}{4} (y - 3 ln ∣ y ∣)

Add a constant to the solution

= \frac{1}{4} (y - 3 ln ∣ y ∣) + C

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What is the integral of (y-3)/(4y) ?
The integral of (y-3)/(4y) is 1/4 (y-3ln|y|)+C