What is the integral from 1 to x of sqrt(t^2-1) ?

The integral from 1 to x of sqrt(t^2-1) is -ln|x+sqrt(x^2-1)|+(xsqrt(x^2-1)+ln|x+sqrt(x^2-1)|)/2

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integral from 1 to x of sqrt(t^2-1)

\int_{1}^{x} t^{2} - 1 d t

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

Solution steps

\int_{1}^{x} t^{2} - 1 d t

Apply Trigonometric Substitution

: \int_{0}^{arcsec (x)} tan^{2} (u) sec (u) d u

= \int_{0}^{arcsec (x)} tan^{2} (u) sec (u) d u

Rewrite using trig identities

= \int_{0}^{arcsec (x)} (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= \int_{0}^{arcsec (x)} - 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_{0}^{arcsec (x)} sec (u) d u + \int_{0}^{arcsec (x)} sec^{3} (u) d u

\int_{0}^{arcsec (x)} sec (u) d u = ln x + x^{2} - 1

\int_{0}^{arcsec (x)} sec^{3} (u) d u = \frac{x x ^{2} - 1 + ln x + x ^{2} - 1}{2}

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

\int_{0}^{2} e^{x} x^{3} d x

\int_{0}^{4} ln (1 + x) d x

\int_{0}^{5} x (5 - x) d x

\int_{\infty}^{0} x e^{- 4 x} d x

\int_{2}^{4} 2 + x d x

What is the integral from 1 to x of sqrt(t^2-1) ?
The integral from 1 to x of sqrt(t^2-1) is -ln|x+sqrt(x^2-1)|+(xsqrt(x^2-1)+ln|x+sqrt(x^2-1)|)/2