What is the integral from 3 to 5 of 1/(sqrt(x^2-9)) ?

The integral from 3 to 5 of 1/(sqrt(x^2-9)) is ln(3)

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integral from 3 to 5 of 1/(sqrt(x^2-9))

\int_{3}^{5} \frac{1}{x ^{2} - 9} d x

ln (3)

Decimal

1.09861 \dots

Solution steps

\int_{3}^{5} \frac{1}{x ^{2} - 9} d x

Apply Trigonometric Substitution

: \int_{0}^{arcsec (\frac{5}{3})} sec (u) d u

= \int_{0}^{arcsec (\frac{5}{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 (\frac{5}{3})}

Compute the boundaries:

ln (3)

= ln (3)

f (x) = (x^{3} - 10 x) (3 x^{2} + 4 x)

y = x^{2} ln (x), y = 4 ln (x)

\frac{d}{d x} ((e^{2 x}) 3 x^{3} + 1)

\frac{d}{d x} (e^{\frac{x + 2}{x - 2}})

\int_{- 8}^{8} (4 x^{9} + 7 x^{5}) d x

What is the integral from 3 to 5 of 1/(sqrt(x^2-9)) ?
The integral from 3 to 5 of 1/(sqrt(x^2-9)) is ln(3)