What is the integral of 1/(sqrt(x^2-6)) ?

The integral of 1/(sqrt(x^2-6)) is ln|sqrt(1/6 (x^2-6))+1/(sqrt(6))x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(sqrt(x^2-6))

\int \frac{1}{x ^{2} - 6} d x

ln \frac{1}{6} (x^{2} - 6) + \frac{1}{6} x + C

Solution steps

\int \frac{1}{x ^{2} - 6} d x

Apply Trigonometric Substitution

= \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 (\frac{1}{6} x)

= ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x))

Simplify

ln tan (arcsec (\frac{1}{6} x)) + sec (arcsec (\frac{1}{6} x)) : ln \frac{1}{6} (x^{2} - 6) + \frac{1}{6} x

= ln \frac{1}{6} (x^{2} - 6) + \frac{1}{6} x

Add a constant to the solution

= ln \frac{1}{6} (x^{2} - 6) + \frac{1}{6} x + C

What is the integral of 1/(sqrt(x^2-6)) ?
The integral of 1/(sqrt(x^2-6)) is ln|sqrt(1/6 (x^2-6))+1/(sqrt(6))x|+C