What is the integral from 4 to 7 of 1/(sqrt(x^2-16)) ?

The integral from 4 to 7 of 1/(sqrt(x^2-16)) is ln(7/4+(sqrt(33))/4)

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integral from 4 to 7 of 1/(sqrt(x^2-16))

\int_{4}^{7} \frac{1}{x ^{2} - 16} d x

ln (\frac{7}{4} + \frac{33}{4})

Solution steps

\int_{4}^{7} \frac{1}{x ^{2} - 16} d x

Apply Trigonometric Substitution

: \int_{0}^{arcsec (\frac{7}{4})} sec (u) d u

= \int_{0}^{arcsec (\frac{7}{4})} 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{7}{4})}

Compute the boundaries:

ln (\frac{7}{4} + \frac{33}{4})

= ln (\frac{7}{4} + \frac{33}{4})

\frac{d}{d x} (2 x)

f (x) = x^{2} - 2 x - 6, at x = - 2

x \to 8 lim (\frac{4}{x - 8})

\int_{0}^{6} 15 4 x + 1 d x

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

What is the integral from 4 to 7 of 1/(sqrt(x^2-16)) ?
The integral from 4 to 7 of 1/(sqrt(x^2-16)) is ln(7/4+(sqrt(33))/4)