What is the integral of (-4y-1)/(y+1) ?

The integral of (-4y-1)/(y+1) is -4y-4+3ln|y+1|+C

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integral of (-4y-1)/(y+1)

\int \frac{- 4 y - 1}{y + 1} d y

- 4 y - 4 + 3 ln ∣ y + 1 ∣ + C

Solution steps

\int \frac{- 4 y - 1}{y + 1} d y

Expand

\frac{- 4 y - 1}{y + 1} : - \frac{4 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{4 y}{y + 1} d y - \int \frac{1}{y + 1} d y

\int \frac{4 y}{y + 1} d y = 4 (y + 1 - ln ∣ y + 1 ∣)

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

= - 4 (y + 1 - ln ∣ y + 1 ∣) - ln ∣ y + 1 ∣

Simplify

- 4 (y + 1 - ln ∣ y + 1 ∣) - ln ∣ y + 1 ∣ : - 4 y - 4 + 3 ln ∣ y + 1 ∣

= - 4 y - 4 + 3 ln ∣ y + 1 ∣

Add a constant to the solution

= - 4 y - 4 + 3 ln ∣ y + 1 ∣ + C

f (x) = x (3 x - 1)

\frac{d}{d x} ((3 x + 4)^{6})

\frac{d}{d x} (- e^{7 x})

\frac{d}{d x} (5 cos (π x))

\int 2 x e^{5 x} d x

What is the integral of (-4y-1)/(y+1) ?
The integral of (-4y-1)/(y+1) is -4y-4+3ln|y+1|+C