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

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

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

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

- ln (3)

Decimal Notation

- 1.09861 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Compute the boundaries:

- ln (3)

= - ln (3)

\int_{1}^{3} ln (x) + sin (x) + 8 d x

\int_{0}^{2} x (x - 2)^{2} d x

\int_{1}^{2} \frac{1}{x ^{3} + 8} d x

\int_{1}^{e} \frac{5}{x} d x

\int_{0}^{1} (x e^{x}) d x

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