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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(u^2-1))

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

ln u^{2} - 1 + u + C

Solution steps

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

v = arcsec (u)

= ln ∣ tan (arcsec (u)) + sec (arcsec (u)) ∣

Simplify

ln ∣ tan (arcsec (u)) + sec (arcsec (u)) ∣ : ln u^{2} - 1 + u

= ln u^{2} - 1 + u

Add a constant to the solution

= ln u^{2} - 1 + u + C

\int_{π}^{0} sin (x) d x

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

\frac{d}{d x} (y + x (ln (x) - ln (y) - 1))

\int 4 x - 5 d x

\frac{s ^{3}}{s + 2}

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