What is the integral of 3/y-1/(y+3) ?

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

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integral of 3/y-1/(y+3)

\int \frac{3}{y} - \frac{1}{y + 3} d y

3 ln ∣ y ∣ - ln ∣ y + 3 ∣ + C

Solution steps

\int \frac{3}{y} - \frac{1}{y + 3} 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 \frac{3}{y} d y - \int \frac{1}{y + 3} d y

\int \frac{3}{y} d y = 3 ln ∣ y ∣

\int \frac{1}{y + 3} d y = ln ∣ y + 3 ∣

= 3 ln ∣ y ∣ - ln ∣ y + 3 ∣

Add a constant to the solution

= 3 ln ∣ y ∣ - ln ∣ y + 3 ∣ + C

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