en

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)

Solution

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

Solution

ln ∣ z - 2 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = z - 2

= ln ∣ z - 2 ∣

Add a constant to the solution

= ln ∣ z - 2 ∣ + C

Graph

Popular Examples

integral of 5/((x-2)^3)

\int \frac{5}{( x - 2 ) ^{3}} d x

integral of cos^4(x/4)

\int cos^{4} (\frac{x}{4}) d x

integral of 3x+4y

\int 3 x + 4 y d x

integral of (x-3)/(x^2+5x+10)

\int \frac{x - 3}{x ^{2} + 5 x + 10} d x

integral of (sqrt(x))/(sqrt(1+\sqrt{x))}

\int \frac{x}{1 + x} d x

Frequently Asked Questions (FAQ)

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