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

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

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

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

ln (\frac{u ^{2} - a ^{2} + u}{∣ a ∣}) + C

Solution steps

\int \frac{1}{u ^{2} - a ^{2}} 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}{a} u)

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

Simplify

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

= ln (\frac{u ^{2} - a ^{2} + u}{∣ a ∣})

Add a constant to the solution

= ln (\frac{u ^{2} - a ^{2} + u}{∣ a ∣}) + C

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