What is the integral of 1/(2y) ?

The integral of 1/(2y) is 1/2 ln|y|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(2y)

\int \frac{1}{2 y} d y

\frac{1}{2} ln ∣ y ∣ + C

Solution steps

\int \frac{1}{2 y} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{y} d y

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= \frac{1}{2} ln ∣ y ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ y ∣ + C

\int_{0}^{8} 5 x + 6 - x d x

t an g e n t y = x^{2} + 8

(arctan (4 x))^{^{'}}

\int \frac{5 x ^{2}}{4 x - x ^{2}} d x

\int \frac{4}{u} d u