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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((1+y))

\int \frac{1}{( 1 + y )} d y

ln ∣ 1 + y ∣ + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 1 + y

= ln ∣ 1 + y ∣

Add a constant to the solution

= ln ∣ 1 + y ∣ + C

\int \frac{1}{1 - cos ^{4} ( x )} d x

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

\int (x + sin (x))^{3} (1 + cos (x)) d x

\int 4 e^{\frac{x}{50}} d x

\int (3 x + 1)^{18} d x