What is the integral of 1/(2y-5) ?

The integral of 1/(2y-5) is 1/2 ln|2y-5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2y-5)

\int \frac{1}{2 y - 5} d y

\frac{1}{2} ln ∣ 2 y - 5 ∣ + C

Solution steps

\int \frac{1}{2 y - 5} d y

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 y - 5

= \frac{1}{2} ln ∣ 2 y - 5 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 y - 5 ∣ + C

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

\int 1 + cos (x) sin (x) d x

\int \frac{1}{( 8 + y ^{2} ) ^{\frac{3}{2}}} d y

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

\int \frac{x - 56}{x ^{2} + x - 56} d x