What is the integral from 2 to 3 of (sqrt(x^2-4))/x ?

The integral from 2 to 3 of (sqrt(x^2-4))/x is -2arcsec(3/2)+sqrt(5)

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integral from 2 to 3 of (sqrt(x^2-4))/x

\int_{2}^{3} \frac{x ^{2} - 4}{x} d x

- 2 arcsec (\frac{3}{2}) + 5

Decimal

0.55393 \dots

Solution steps

\int_{2}^{3} \frac{x ^{2} - 4}{x} d x

Apply Trigonometric Substitution

: \int_{0}^{arcsec (\frac{3}{2})} 2 tan^{2} (u) d u

= \int_{0}^{arcsec (\frac{3}{2})} 2 tan^{2} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{arcsec (\frac{3}{2})} tan^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \int_{0}^{arcsec (\frac{3}{2})} - 1 + sec^{2} (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

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

\int_{0}^{arcsec (\frac{3}{2})} 1 d u = arcsec (\frac{3}{2})

\int_{0}^{arcsec (\frac{3}{2})} sec^{2} (u) d u = \frac{5}{2}

= 2 (- arcsec (\frac{3}{2}) + \frac{5}{2})

Simplify

2 (- arcsec (\frac{3}{2}) + \frac{5}{2}) : - 2 arcsec (\frac{3}{2}) + 5

= - 2 arcsec (\frac{3}{2}) + 5

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What is the integral from 2 to 3 of (sqrt(x^2-4))/x ?
The integral from 2 to 3 of (sqrt(x^2-4))/x is -2arcsec(3/2)+sqrt(5)