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

The integral of 1/(sqrt((x^2-16))) is ln(1/4 |sqrt(x^2-16)+x|)+C

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integral of 1/(sqrt((x^2-16)))

\int \frac{1}{( x ^{2} - 16 )} d x

ln (\frac{1}{4} x^{2} - 16 + x) + C

Solution steps

\int \frac{1}{x ^{2} - 16} 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}{4} x)

= ln tan (arcsec (\frac{1}{4} x)) + sec (arcsec (\frac{1}{4} x))

Simplify

ln tan (arcsec (\frac{1}{4} x)) + sec (arcsec (\frac{1}{4} x)) : ln (\frac{1}{4} x^{2} - 16 + x)

= ln (\frac{1}{4} x^{2} - 16 + x)

Add a constant to the solution

= ln (\frac{1}{4} x^{2} - 16 + x) + C

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What is the integral of 1/(sqrt((x^2-16))) ?
The integral of 1/(sqrt((x^2-16))) is ln(1/4 |sqrt(x^2-16)+x|)+C