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

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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

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

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

d er i v a t i v e f (x) = (x - x^{2})

(y) d x + (y - x) d y = 0

\frac{d}{d x} (a n x + b 3 x)

t an g e n t y = \frac{5}{4 x + 1}, (1, 1)

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