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

\frac{d}{d x} (3 x ln (4 x) - 3 x)

\frac{\partial}{\partial x} (6 ze x yz)

t y^{^{'}} + 3 y = t^{2}

\int 18 θ sec^{2} (θ) d θ

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