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

The integral of (y-1)/(y+4) is -5ln|y+4|+y+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 5 ln ∣ y + 4 ∣ + y + C

Solution steps

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

\frac{y - 1}{y + 4} = - \frac{5}{y + 4} + 1

= \int - \frac{5}{y + 4} + 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{5}{y + 4} d y + \int 1 d y

\int \frac{5}{y + 4} d y = 5 ln ∣ y + 4 ∣

\int 1 d y = y

= - 5 ln ∣ y + 4 ∣ + y

Add a constant to the solution

= - 5 ln ∣ y + 4 ∣ + y + C

x \to 3 lim (6 x^{2})

\int 8 sec^{3} (6 x) d x

\frac{\partial}{\partial x} (3 e^{2 x} - 5 ln (y) + 7)

\int_{2}^{7} x d x

\frac{d y}{d x} = (\frac{3 y + 2}{5 x + 4})^{2}

What is the integral of (y-1)/(y+4) ?
The integral of (y-1)/(y+4) is -5ln|y+4|+y+C