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

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

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integral of sqrt(u^2-1)

\int u^{2} - 1 d u

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

Solution steps

\int u^{2} - 1 d u

Apply Trigonometric Substitution

= \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= \int - sec (v) + sec^{3} (v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int sec (v) d v + \int sec^{3} (v) d v

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

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= - ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

v = arcsec (u)

= - ln ∣ tan (arcsec (u)) + sec (arcsec (u)) ∣ + \frac{1}{2} sec (arcsec (u)) tan (arcsec (u)) + \frac{1}{2} ln ∣ tan (arcsec (u)) + sec (arcsec (u)) ∣

Simplify

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

= \frac{1}{2} u u^{2} - 1 - \frac{1}{2} ln u + u^{2} - 1

Add a constant to the solution

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

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