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

Take the constant out:

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

= - \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

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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