What is the integral of 1/(10-x) ?

The integral of 1/(10-x) is -ln|10-x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(10-x)

\int \frac{1}{10 - x} d x

- ln ∣ 10 - x ∣ + C

Solution steps

\int \frac{1}{10 - x} d x

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 = 10 - x

= - ln ∣ 10 - x ∣

Add a constant to the solution

= - ln ∣ 10 - x ∣ + C

\frac{\partial}{\partial z} (x tan (y) - z e^{z})

\frac{d y}{d t} + 3 y = t + e^{- 2 t}

\int \frac{4}{3 2 x + 1} d x

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

\frac{d}{d t} (csc (t))