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

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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

= \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

n = 1 \sum \infty 5

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\frac{9}{x ^{2}} - \frac{3}{x}

\int 2 e^{x} \cdot e^{2 x} d 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