What is the integral from 3 to 5 of 1/(sqrt(x^2-4)) ?

The integral from 3 to 5 of 1/(sqrt(x^2-4)) is ln(-((sqrt(21)+5)(sqrt(5)-3))/4)

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integral from 3 to 5 of 1/(sqrt(x^2-4))

\int_{3}^{5} \frac{1}{x ^{2} - 4} d x

ln (- \frac{( 21 + 5 ) ( 5 - 3 )}{4})

Solution steps

\int_{3}^{5} \frac{1}{x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int_{arcsec (\frac{3}{2})}^{arcsec (\frac{5}{2})} sec (u) d u

= \int_{arcsec (\frac{3}{2})}^{arcsec (\frac{5}{2})} sec (u) d u

Use the common integral:

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

= [ln ∣ tan (u) + sec (u) ∣]_{arcsec (\frac{3}{2})}^{arcsec (\frac{5}{2})}

Compute the boundaries:

ln (\frac{21}{2} + \frac{5}{2}) - ln (\frac{5}{2} + \frac{3}{2})

= ln (\frac{21}{2} + \frac{5}{2}) - ln (\frac{5}{2} + \frac{3}{2})

Simplify

= ln (- \frac{( 21 + 5 ) ( 5 - 3 )}{4})

\int_{1}^{e} \frac{( ln ( x ) )}{x ^{2}} d x

\int_{0}^{x} x + 2 y d y

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

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

\int_{- 5}^{- 3} - x^{2} - 3 x d x

What is the integral from 3 to 5 of 1/(sqrt(x^2-4)) ?
The integral from 3 to 5 of 1/(sqrt(x^2-4)) is ln(-((sqrt(21)+5)(sqrt(5)-3))/4)