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

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

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

\int \frac{1}{u ^{2} - 16} d u

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

Solution steps

\int \frac{1}{u ^{2} - 16} d u

Apply Trigonometric Substitution

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

v = arcsec (\frac{1}{4} u)

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

Simplify

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

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

Add a constant to the solution

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

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