What is the integral of 1/(sqrt(x^2-6x+8)) ?

The integral of 1/(sqrt(x^2-6x+8)) is ln|sqrt(x^2-6x+8)+x-3|+C

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integral of 1/(sqrt(x^2-6x+8))

\int \frac{1}{x ^{2} - 6 x + 8} d x

ln x^{2} - 6 x + 8 + x - 3 + C

Solution steps

\int \frac{1}{x ^{2} - 6 x + 8} d x

Complete the square

x^{2} - 6 x + 8 : (x - 3)^{2} - 1

= \int \frac{1}{( x - 3 ) ^{2} - 1} d x

Apply u-substitution

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

= \int \frac{1}{u ^{2} - 1} 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 (x - 3)) + sec (arcsec (x - 3)) ∣

Simplify

ln ∣ tan (arcsec (x - 3)) + sec (arcsec (x - 3)) ∣ : ln x^{2} - 6 x + 8 + x - 3

= ln x^{2} - 6 x + 8 + x - 3

Add a constant to the solution

= ln x^{2} - 6 x + 8 + x - 3 + C

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What is the integral of 1/(sqrt(x^2-6x+8)) ?
The integral of 1/(sqrt(x^2-6x+8)) is ln|sqrt(x^2-6x+8)+x-3|+C