What is the integral of 1/(sqrt(x^2-2x)) ?

The integral of 1/(sqrt(x^2-2x)) is ln|sqrt(x^2-2x)+x-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/(sqrt(x^2-2x))

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

ln x^{2} - 2 x + x - 1 + C

Solution steps

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

Complete the square

x^{2} - 2 x : (x - 1)^{2} - 1

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 1} d u

= \int \frac{1}{u ^{2} - 1} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln ∣ tan (arcsec (x - 1)) + sec (arcsec (x - 1)) ∣

Simplify

ln ∣ tan (arcsec (x - 1)) + sec (arcsec (x - 1)) ∣ : ln x^{2} - 2 x + x - 1

= ln x^{2} - 2 x + x - 1

Add a constant to the solution

= ln x^{2} - 2 x + x - 1 + C

3 x^{2}, (1, 3)

\int_{0}^{π} 12 sin^{4} (x) d x

\int_{1}^{2} (7 x + x^{2}) d x

cos (2^{x})

(- 19 x y^{2} + 10 y) + (- 19 x^{2} y + 10 x) y^{^{'}} = 0

What is the integral of 1/(sqrt(x^2-2x)) ?
The integral of 1/(sqrt(x^2-2x)) is ln|sqrt(x^2-2x)+x-1|+C