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

The integral from 2 to 3 of 1/(sqrt(x^2-1)) is ln(2sqrt(2)+3)-ln(sqrt(3)+2)

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

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

ln (22 + 3) - ln (3 + 2)

Decimal

0.44578 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{\frac{π}{3}}^{arcsec (3)} sec (u) d u

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

Use the common integral:

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

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

Compute the boundaries:

ln (22 + 3) - ln (3 + 2)

= ln (22 + 3) - ln (3 + 2)

\int_{- 1}^{1} - x^{2} ∣ x ∣ d x

\int_{0}^{10} \frac{1}{10} x d x

\int_{- 1}^{1} (1 + x^{4} tan (x)) d x

\int_{0}^{π} x^{3} d x

\int_{0}^{0.5} 1 - x^{2} d x

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