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

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

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

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

\frac{y ^{2}}{- y - 1} + 2 ln ∣ y + 1 ∣ - \frac{1}{- y - 1} + C

Solution steps

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

Expand

\frac{y - 1}{- y - 1} : \frac{y}{- y - 1} - \frac{1}{- y - 1}

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

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

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

= \frac{y ^{2}}{- y - 1} + ln ∣ y + 1 ∣ - \frac{1}{- y - 1} - (- ln ∣ y + 1 ∣)

Simplify

= \frac{y ^{2}}{- y - 1} + 2 ln ∣ y + 1 ∣ - \frac{1}{- y - 1}

Add a constant to the solution

= \frac{y ^{2}}{- y - 1} + 2 ln ∣ y + 1 ∣ - \frac{1}{- y - 1} + C

\frac{\partial}{\partial y} (2 (y + 1) e^{2 x})

(- 6, 3), (4, - 1)

- 19.6 x + 2

\int \frac{1}{x + 1} d x

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