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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

= \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}{a} x)

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

Simplify

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

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

Add a constant to the solution

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

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