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

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

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

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

ln (2 + 3)

Decimal Notation

1.31695 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{3}} sec (u) d u

= \int_{0}^{\frac{π}{3}} sec (u) d u

Use the common integral:

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

= [ln ∣ tan (u) + sec (u) ∣]_{0}^{\frac{π}{3}}

Compute the boundaries:

ln (2 + 3)

= ln (2 + 3)

\int_{1}^{e} \frac{1}{x ( 1 + ln ( x ) )} d x

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

\int_{0}^{s i n (x)} ln (2 + t^{2}) d t

\int_{0}^{5} ∣ 2 x - 4 ∣ d x

\int_{\frac{π}{6}}^{\frac{π}{2}} cos (x) d x

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