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

The integral of 1/(2t-1) is 1/2 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 1/(2t-1)

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

\frac{1}{2} ln ∣ 2 t - 1 ∣ + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 t - 1

= \frac{1}{2} ln ∣ 2 t - 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 t - 1 ∣ + C

\int (4 x - 5)^{\frac{2}{3}} d x

\int (1 + 6 x^{2} - 10 x^{4}) d x

\int \frac{( 2 x - 7 )}{( x ^{(2)} + 9 )} d x

\int 6 sec (x) d x

\int \frac{x - 4}{x ^{2} + 4 x + 3} d x