What is the integral of 1/(sqrt(x^2-4)) ?

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

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integral of 1/(sqrt(x^2-4))

\int \frac{1}{x ^{2} - 4} d x

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

Solution steps

\int \frac{1}{x ^{2} - 4} 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}{2} x)

= ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{2}{x}

\frac{d}{d x} (\frac{a}{1 + \frac{b}{x + c}})

\int 6 x^{2} d x

x \to 0 lim (x - \frac{1}{ln ∣ x ∣})

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 6

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