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

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

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integral of 1/(sqrt(x^2-4x+8))

\int \frac{1}{x ^{2} - 4 x + 8} d x

ln (\frac{1}{2} x - 2 + x^{2} - 4 x + 8) + C

Solution steps

\int \frac{1}{x ^{2} - 4 x + 8} d x

Complete the square

x^{2} - 4 x + 8 : (x - 2)^{2} + 4

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= \int \frac{1}{u ^{2} + 4} 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 (arctan (\frac{1}{2} (x - 2))) + sec (arctan (\frac{1}{2} (x - 2)))

Simplify

ln tan (arctan (\frac{1}{2} (x - 2))) + sec (arctan (\frac{1}{2} (x - 2))) : ln (\frac{1}{2} x - 2 + x^{2} - 4 x + 8)

= ln (\frac{1}{2} x - 2 + x^{2} - 4 x + 8)

Add a constant to the solution

= ln (\frac{1}{2} x - 2 + x^{2} - 4 x + 8) + C

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