What is the integral of 1/((x-3)sqrt(x^2-6x+6)) ?

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

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integral of 1/((x-3)sqrt(x^2-6x+6))

\int \frac{1}{( x - 3 ) x ^{2} - 6 x + 6} d x

\frac{1}{3} arcsec (\frac{1}{3} (x - 3)) + C

Solution steps

\int \frac{1}{( x - 3 ) x ^{2} - 6 x + 6} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 3} d u

= \int \frac{1}{u u ^{2} - 3} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (x - 3))

Simplify

= \frac{1}{3} arcsec (\frac{1}{3} (x - 3))

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{1}{3} (x - 3)) + C

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