What is the integral from 4 to 5 of 1/(sqrt(x^2-16)) ?

The integral from 4 to 5 of 1/(sqrt(x^2-16)) is ln(2)

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integral from 4 to 5 of 1/(sqrt(x^2-16))

\int_{4}^{5} \frac{1}{x ^{2} - 16} d x

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{4}^{5} \frac{1}{x ^{2} - 16} d x

Apply Trigonometric Substitution

: \int_{0}^{arcsec (\frac{5}{4})} sec (u) d u

= \int_{0}^{arcsec (\frac{5}{4})} sec (u) d u

Use the common integral:

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

= [ln ∣ tan (u) + sec (u) ∣]_{0}^{arcsec (\frac{5}{4})}

Compute the boundaries:

ln (2)

= ln (2)

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What is the integral from 4 to 5 of 1/(sqrt(x^2-16)) ?
The integral from 4 to 5 of 1/(sqrt(x^2-16)) is ln(2)