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

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

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

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

ln \frac{1}{6} (x^{2} + 4 x - 2) + \frac{1}{6} (x + 2) + C

Solution steps

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

Complete the square

x^{2} + 4 x - 2 : (x + 2)^{2} - 6

= \int \frac{1}{( x + 2 ) ^{2} - 6} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 6} d u

= \int \frac{1}{u ^{2} - 6} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arcsec (\frac{1}{6} (x + 2))) + sec (arcsec (\frac{1}{6} (x + 2)))

Simplify

ln tan (arcsec (\frac{1}{6} (x + 2))) + sec (arcsec (\frac{1}{6} (x + 2))) : ln \frac{1}{6} (x^{2} + 4 x - 2) + \frac{1}{6} (x + 2)

= ln \frac{1}{6} (x^{2} + 4 x - 2) + \frac{1}{6} (x + 2)

Add a constant to the solution

= ln \frac{1}{6} (x^{2} + 4 x - 2) + \frac{1}{6} (x + 2) + C

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