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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

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

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (x)

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

Simplify

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

= ln x^{2} - 1 + x

Add a constant to the solution

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

ln (1 + x), 5

2 x^{2} - 4, x^{2}

\frac{d}{d x} (2 e^{x} (cos (x) - sin (x)))

f (x) = (x - x^{2})

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