What is the integral of 1/(sqrt(2x^2-8)) ?

The integral of 1/(sqrt(2x^2-8)) is 1/(sqrt(2))ln((|sqrt(x^2-4)+x|)/2)+C

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integral of 1/(sqrt(2x^2-8))

\int \frac{1}{2 x ^{2} - 8} d x

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

Solution steps

\int \frac{1}{2 x ^{2} - 8} d x

\frac{1}{2 x ^{2} - 8} = \frac{1}{2} \frac{1}{x ^{2} - 4}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

= \frac{1}{2} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{1}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of 1/(sqrt(2x^2-8)) ?
The integral of 1/(sqrt(2x^2-8)) is 1/(sqrt(2))ln((|sqrt(x^2-4)+x|)/2)+C