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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

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 (t)

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

Simplify

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

= ln t^{2} - 1 + t

Add a constant to the solution

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

\int (x^{3} + 1) e^{x^{4} + 4 x} d x

\int \frac{8 x ^{4} - 9 x ^{3} + 6 x ^{2} - 2 x + 1}{x ^{2}} d x

\int (5 x^{2} + 2)^{6} 10 d x

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

\int \frac{1}{5 t ^{4}} d t

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