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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 5 ln ∣ u + 4 ∣ + u + C

Solution steps

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

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

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

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}{u + 4} d u + \int 1 d u

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

\int 1 d u = u

= - 5 ln ∣ u + 4 ∣ + u

Add a constant to the solution

= - 5 ln ∣ u + 4 ∣ + u + C

\int tan (4 x) sec (4 x) d x

\int - \frac{2 z}{5} d z

\int 14 x ln (8 x) d x

\int \frac{6 x}{3 x ^{2}} d x

\int (4 x^{3} + 4 x) e^{x^{4} + 2 x^{2} + 1} d x

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