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

The integral of (y+4)/(y-4) is 8ln|y-4|+y+C

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

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

8 ln ∣ y - 4 ∣ + y + C

Solution steps

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

\frac{y + 4}{y - 4} = \frac{8}{y - 4} + 1

= \int \frac{8}{y - 4} + 1 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{8}{y - 4} d y + \int 1 d y

\int \frac{8}{y - 4} d y = 8 ln ∣ y - 4 ∣

\int 1 d y = y

= 8 ln ∣ y - 4 ∣ + y

Add a constant to the solution

= 8 ln ∣ y - 4 ∣ + y + C

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What is the integral of (y+4)/(y-4) ?
The integral of (y+4)/(y-4) is 8ln|y-4|+y+C