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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(z^2-1)

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

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

Solution steps

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

Factor

z^{2} - 1 : - (- z^{2} + 1)

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = (6 x^{2} - 10 x) e^{x}

\int (sec^{2} (θ) + 5) d θ

\frac{d y}{d x} + y^{9} x + 7 y = 0

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

\frac{\partial}{\partial x} (2 x^{2} y - x^{3})

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