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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2/(2t-1)

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

ln ∣ 2 t - 1 ∣ + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{1}{2 t - 1} d t

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 t - 1

= 2 \cdot \frac{1}{2} ln ∣ 2 t - 1 ∣

Simplify

2 \cdot \frac{1}{2} ln ∣ 2 t - 1 ∣ : ln ∣ 2 t - 1 ∣

= ln ∣ 2 t - 1 ∣

Add a constant to the solution

= ln ∣ 2 t - 1 ∣ + C