What is the integral of 1/(sqrt(x^2-5)) ?

The integral of 1/(sqrt(x^2-5)) is ln|sqrt(1/5 (x^2-5))+1/(sqrt(5))x|+C

integral of 1/(sqrt(x^2-5))

\int \frac{1}{x ^{2} - 5} d x

ln \frac{1}{5} (x^{2} - 5) + \frac{1}{5} x + C

Solution steps

\int \frac{1}{x ^{2} - 5} 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}{5} x)

= ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x))

Simplify

ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)) : ln \frac{1}{5} (x^{2} - 5) + \frac{1}{5} x

= ln \frac{1}{5} (x^{2} - 5) + \frac{1}{5} x

Add a constant to the solution

= ln \frac{1}{5} (x^{2} - 5) + \frac{1}{5} x + C

\int - 12 x d x

\int 2 t^{3} (t^{4} - 3)^{5} d t

g (x) = \frac{arcsin ( 3 x )}{x}

x \to - \infty lim (- 3 x^{3} + x)

x \to 0 lim (- \frac{ln ( 1 + 6 x )}{x})

What is the integral of 1/(sqrt(x^2-5)) ?
The integral of 1/(sqrt(x^2-5)) is ln|sqrt(1/5 (x^2-5))+1/(sqrt(5))x|+C