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

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

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

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

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

Decimal Notation

1.56679 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{arcsec (\frac{5}{2})} sec (u) d u

= \int_{0}^{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) ∣]_{0}^{arcsec (\frac{5}{2})}

Compute the boundaries:

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

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

\int \frac{x ^{5}}{( x ^{2} + 4 ) ^{2}} d x

\frac{d}{d x} (6 x^{3} - 4 x^{2} + 5 x)

f (x) = \frac{9 x}{x + 4}, at x = 8

\int - \frac{1}{4} sin (2 x) d x

f (x) = \frac{6 x}{x + 6}

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