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

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

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integral of sqrt(y^2-1)

\int y^{2} - 1 d y

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

Solution steps

\int y^{2} - 1 d y

Apply Trigonometric Substitution

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

= \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 (y)

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{3} - 2 x + 2, (4, 58)

e^{- t} (cos (4 t) - 2 sin (4 t))

\frac{d}{d t} (t^{2} e^{t})

\frac{3}{( x ^{2} - x - 2 )}

\frac{d}{d x} ((2 + x)^{\frac{1}{3}} (1 - x)^{\frac{2}{3}})

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