What is the integral of (y-3)/(y-2) ?

The integral of (y-3)/(y-2) is -ln|y-2|+y+C

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integral of (y-3)/(y-2)

\int \frac{y - 3}{y - 2} d y

- ln ∣ y - 2 ∣ + y + C

Solution steps

\int \frac{y - 3}{y - 2} d y

\frac{y - 3}{y - 2} = - \frac{1}{y - 2} + 1

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

\int \frac{1}{y - 2} d y = ln ∣ y - 2 ∣

\int 1 d y = y

= - ln ∣ y - 2 ∣ + y

Add a constant to the solution

= - ln ∣ y - 2 ∣ + y + C

\int e^{x + 4} d x

y^{^{'}} = (x + y)^{2} - 1

\frac{2}{s ^{2} ( s ^{2} + 3 s + 10 )}

tan (f (x)) (x) = - 4 x + \frac{1}{2} x - 5, at x = 1

\frac{\partial}{\partial x} (3 x^{2 y})

What is the integral of (y-3)/(y-2) ?
The integral of (y-3)/(y-2) is -ln|y-2|+y+C