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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (3 x)

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

Simplify

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

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

Add a constant to the solution

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

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