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

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

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

\frac{1}{10} ln ∣ 10 x - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{10 u} d u

= \int \frac{1}{10 u} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{10} ln ∣ u ∣

Substitute back

u = 10 x - 1

= \frac{1}{10} ln ∣ 10 x - 1 ∣

Add a constant to the solution

= \frac{1}{10} ln ∣ 10 x - 1 ∣ + C

\int e^{i w x} d x

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

\int \frac{2}{50 + t} d t

\int \frac{1}{( x 1 - x ^{2} )} d x

\int x^{1.8} + 11 x^{4.5} d x

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