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

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

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

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

ln (22 + 3)

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{arcsec (3)} sec (u) d u

= \int_{0}^{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) ∣]_{0}^{arcsec (3)}

Compute the boundaries:

ln (22 + 3)

= ln (22 + 3)

\int_{1}^{4} (x^{2} - 4 x + 5) d x

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

\int_{π}^{2 π} r sin (r) d r

\int_{\frac{- π}{2}}^{\frac{π}{2}} g (x) d x

\int_{0}^{1} (\frac{x - 1}{x + 1})^{3} d x

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