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

The integral from 1 to 5 of 1/(sqrt(x^2-1)) is ln(2sqrt(6)+5)

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

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

ln (26 + 5)

Decimal Notation

2.29243 \dots

Solution steps

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

Apply Trigonometric Substitution

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

= \int_{0}^{arcsec (5)} 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 (5)}

Compute the boundaries:

ln (26 + 5)

= ln (26 + 5)

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What is the integral from 1 to 5 of 1/(sqrt(x^2-1)) ?
The integral from 1 to 5 of 1/(sqrt(x^2-1)) is ln(2sqrt(6)+5)