What is the integral of (x^2-1)^{0.5} ?

The integral of (x^2-1)^{0.5} is 1/2 xsqrt(x^2-1)-1/2 ln|x+sqrt(x^2-1)|+C

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integral of (x^2-1)^{0.5}

\int (x^{2} - 1)^{0.5} d x

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

Solution steps

\int (x^{2} - 1)^{0.5} d x

Simplify

x^{2} : xx

= \int (xx - 1)^{0.5} d x

xx = x^{2}

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

Apply Trigonometric Substitution

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

Rewrite using trig identities

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

Expand

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

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

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 (u) d u + \int sec^{3} (u) d u

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

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

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

Substitute back

u = arcsec (x)

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of (x^2-1)^{0.5} ?
The integral of (x^2-1)^{0.5} is 1/2 xsqrt(x^2-1)-1/2 ln|x+sqrt(x^2-1)|+C