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

= \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

\int 1 - 3 x d x

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

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

\int (x^{3} + 6 x) (6 x^{2} + 12) d x

\int - tan (x) sec (x) d x

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