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

Popular >

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

: \int \frac{1}{2 u} d u

= 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

\int \frac{2 + ln ( x )}{x} d x

\frac{d}{d x} (ln (1 + \frac{2}{x}))

\int_{- 1}^{5} \frac{8 y}{y ^{2} - 4 y - 12} d y

\int 16 csc (8 x) sec (8 x) d x

x^{2} + 2 x + 1, - 1, 7