What is the integral of 1/(sqrt(2x^2-4x-6)) ?

The integral of 1/(sqrt(2x^2-4x-6)) is 1/(sqrt(2))ln((|sqrt(x^2-2x-3)+x-1|)/2)+C

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integral of 1/(sqrt(2x^2-4x-6))

\int \frac{1}{2 x ^{2} - 4 x - 6} d x

\frac{1}{2} ln (\frac{x ^{2} - 2 x - 3 + x - 1}{2}) + C

Solution steps

\int \frac{1}{2 x ^{2} - 4 x - 6} d x

Complete the square

2 x^{2} - 4 x - 6 : 2 (x - 1)^{2} - 8

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

Apply u-substitution

: \int \frac{1}{2 u ^{2} - 8} d u

= \int \frac{1}{2 u ^{2} - 8} d u

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

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

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}{u ^{2} - 4} d u

Apply Trigonometric Substitution

: \int sec (v) d v

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

Use the common integral:

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

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

Substitute back

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

Simplify

\frac{1}{2} ln tan (arcsec (\frac{1}{2} (x - 1))) + sec (arcsec (\frac{1}{2} (x - 1))) : \frac{1}{2} ln (\frac{x ^{2} - 2 x - 3 + x - 1}{2})

= \frac{1}{2} ln (\frac{x ^{2} - 2 x - 3 + x - 1}{2})

Add a constant to the solution

= \frac{1}{2} ln (\frac{x ^{2} - 2 x - 3 + x - 1}{2}) + C

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What is the integral of 1/(sqrt(2x^2-4x-6)) ?
The integral of 1/(sqrt(2x^2-4x-6)) is 1/(sqrt(2))ln((|sqrt(x^2-2x-3)+x-1|)/2)+C