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

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

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integral of (-2y-2)/(y^2+y)

\int \frac{- 2 y - 2}{y ^{2} + y} d y

- 2 ln ∣ y ∣ + C

Solution steps

\int \frac{- 2 y - 2}{y ^{2} + y} d y

Expand

\frac{- 2 y - 2}{y ^{2} + y} : - \frac{2 y}{y ^{2} + y} - \frac{2}{y ^{2} + 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}{y ^{2} + y} d y - \int \frac{2}{y ^{2} + y} d y

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

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

= - 2 ln ∣ y + 1 ∣ - 2 (ln ∣ y ∣ - ln ∣ y + 1 ∣)

Simplify

- 2 ln ∣ y + 1 ∣ - 2 (ln ∣ y ∣ - ln ∣ y + 1 ∣) : - 2 ln ∣ y ∣

= - 2 ln ∣ y ∣

Add a constant to the solution

= - 2 ln ∣ y ∣ + C

x \to 0 - lim (\frac{x}{∣ x + 1 ∣})

\int \frac{16 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

\int (\frac{1}{8}) y_{1} e^{- \frac{y _{1} + y _{2}}{2}} d y_{1}

f (x) = 3 sin (x) - 4 cos (x)

y^{^{'}} = 4 x y

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