What is the integral from 3 to 8 of 1/(sqrt(x^2-4)) ?

The integral from 3 to 8 of 1/(sqrt(x^2-4)) is ln((3sqrt(15)+12-5sqrt(3)-4sqrt(5))/2)

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integral from 3 to 8 of 1/(sqrt(x^2-4))

\int_{3}^{8} \frac{1}{x ^{2} - 4} d x

ln (\frac{3 15 + 12 - 5 3 - 4 5}{2})

Decimal

1.10101 \dots

Solution steps

\int_{3}^{8} \frac{1}{x ^{2} - 4} d x

Apply Trigonometric Substitution

= \int_{arcsec (\frac{3}{2})}^{arcsec (4)} sec (u) d u

Use the common integral:

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

= [ln ∣ tan (u) + sec (u) ∣]_{arcsec (\frac{3}{2})}^{arcsec (4)}

Compute the boundaries:

ln (15 + 4) - ln (\frac{5}{2} + \frac{3}{2})

= ln (15 + 4) - ln (\frac{5}{2} + \frac{3}{2})

Simplify

= ln (\frac{3 15 + 12 - 5 3 - 4 5}{2})

What is the integral from 3 to 8 of 1/(sqrt(x^2-4)) ?
The integral from 3 to 8 of 1/(sqrt(x^2-4)) is ln((3sqrt(15)+12-5sqrt(3)-4sqrt(5))/2)