What is the integral of (1-z)/(z^2+2z+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (1-z)/(z^2+2z+1)

\int \frac{1 - z}{z ^{2} + 2 z + 1} d z

- \frac{2}{z + 1} - ln ∣ z + 1 ∣ + C

Solution steps

\int \frac{1 - z}{z ^{2} + 2 z + 1} d z

Expand

\frac{1 - z}{z ^{2} + 2 z + 1} : \frac{1}{z ^{2} + 2 z + 1} - \frac{z}{z ^{2} + 2 z + 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{1}{z ^{2} + 2 z + 1} d z - \int \frac{z}{z ^{2} + 2 z + 1} d z

\int \frac{1}{z ^{2} + 2 z + 1} d z = - \frac{1}{z + 1}

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

= - \frac{1}{z + 1} - (ln ∣ z + 1 ∣ + \frac{1}{z + 1})

Simplify

- \frac{1}{z + 1} - (ln ∣ z + 1 ∣ + \frac{1}{z + 1}) : - \frac{2}{z + 1} - ln ∣ z + 1 ∣

= - \frac{2}{z + 1} - ln ∣ z + 1 ∣

Add a constant to the solution

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

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