What is the integral of 1/(sqrt(x^2-64)) ?

The integral of 1/(sqrt(x^2-64)) is ln(1/8 |sqrt(x^2-64)+x|)+C

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integral of 1/(sqrt(x^2-64))

\int \frac{1}{x ^{2} - 64} d x

ln (\frac{1}{8} x^{2} - 64 + x) + C

Solution steps

\int \frac{1}{x ^{2} - 64} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

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

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{1}{8} x)

= ln tan (arcsec (\frac{1}{8} x)) + sec (arcsec (\frac{1}{8} x))

Simplify

ln tan (arcsec (\frac{1}{8} x)) + sec (arcsec (\frac{1}{8} x)) : ln (\frac{1}{8} x^{2} - 64 + x)

= ln (\frac{1}{8} x^{2} - 64 + x)

Add a constant to the solution

= ln (\frac{1}{8} x^{2} - 64 + x) + C

x \to - 2 lim (8 + 1^{3})

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

\frac{d y}{d x} = \frac{x + y}{y - x}

f (x) = 10 x^{3} - 45 x^{2} + 60 x - 30

\frac{d}{d s} (\frac{( s ^{2} - 4 s + 20 )}{s ^{2} + 6 s + 8})

What is the integral of 1/(sqrt(x^2-64)) ?
The integral of 1/(sqrt(x^2-64)) is ln(1/8 |sqrt(x^2-64)+x|)+C