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

Popular >

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

\frac{\partial}{\partial x} (3 x^{2} y + 4 x y^{2})

x \frac{d y}{d x} - 2 y = 2

\frac{\partial}{\partial y} (x ln (x^{2} + y^{2}))

\frac{d}{d x} (400 (15 - x^{2}) (3 x - 2))

\int cos (b 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