What is the integral of 1/(20-y) ?

The integral of 1/(20-y) is -ln|20-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/(20-y)

\int \frac{1}{20 - y} d y

- ln ∣ 20 - y ∣ + C

Solution steps

\int \frac{1}{20 - y} d y

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 20 - y

= - ln ∣ 20 - y ∣

Add a constant to the solution

= - ln ∣ 20 - y ∣ + C

\frac{d}{d x} (sin (ln (x)) + cos (ln (x)))

f (x) = - 3 x^{3}

\frac{( s + 6 )}{( s ^{2} + 6 s + 25 )}

\frac{d}{d x} (x + \frac{4}{x})

\int (x + 1)^{2} ln (x + 1) d x