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

The integral of (y+1)/(y-1) is 2ln|y-1|+y+C

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

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

2 ln ∣ y - 1 ∣ + y + C

Solution steps

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

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

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

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

\int 1 d y = y

= 2 ln ∣ y - 1 ∣ + y

Add a constant to the solution

= 2 ln ∣ y - 1 ∣ + y + C

y^{^{''}} - 5 y^{^{'}} = x - 2

3 x^{2} cos (x) cot (x)

e^{x} sin (2 x)

\frac{d}{d x} (ln (\frac{x}{k}))

\int_{- a}^{a} \frac{1}{( s - x ) ^{2}} d x

What is the integral of (y+1)/(y-1) ?
The integral of (y+1)/(y-1) is 2ln|y-1|+y+C