What is the integral of 1/(sqrt(x^2-3)) ?

The integral of 1/(sqrt(x^2-3)) is ln|sqrt(1/3 (x^2-3))+1/(sqrt(3))x|+C

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integral of 1/(sqrt(x^2-3))

\int \frac{1}{x ^{2} - 3} d x

ln \frac{1}{3} (x^{2} - 3) + \frac{1}{3} x + C

Solution steps

\int \frac{1}{x ^{2} - 3} 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 (\frac{1}{3} x)

= ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x))

Simplify

ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x)) : ln \frac{1}{3} (x^{2} - 3) + \frac{1}{3} x

= ln \frac{1}{3} (x^{2} - 3) + \frac{1}{3} x

Add a constant to the solution

= ln \frac{1}{3} (x^{2} - 3) + \frac{1}{3} x + C

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What is the integral of 1/(sqrt(x^2-3)) ?
The integral of 1/(sqrt(x^2-3)) is ln|sqrt(1/3 (x^2-3))+1/(sqrt(3))x|+C