What is the integral of 1/((x-6)sqrt(x^2-12x)) ?

The integral of 1/((x-6)sqrt(x^2-12x)) is 1/6 arcsec(1/6 x-1)+C

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integral of 1/((x-6)sqrt(x^2-12x))

\int \frac{1}{( x - 6 ) x ^{2} - 12 x} d x

\frac{1}{6} arcsec (\frac{1}{6} x - 1) + C

Solution steps

\int \frac{1}{( x - 6 ) x ^{2} - 12 x} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 36} d u

= \int \frac{1}{u u ^{2} - 36} d u

Apply Trigonometric Substitution

: \int \frac{1}{6} d v

= \int \frac{1}{6} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{6} \cdot 1 \cdot v

Substitute back

= \frac{1}{6} \cdot 1 \cdot arcsec (\frac{1}{6} (x - 6))

Simplify

\frac{1}{6} \cdot 1 \cdot arcsec (\frac{1}{6} (x - 6)) : \frac{1}{6} arcsec (\frac{1}{6} x - 1)

= \frac{1}{6} arcsec (\frac{1}{6} x - 1)

Add a constant to the solution

= \frac{1}{6} arcsec (\frac{1}{6} x - 1) + C

\frac{d y}{d x} + (2 x y^{2}) = 0

\int_{1}^{8} x^{- \frac{5}{3}} d x

e^{x} (7 - 5 x + 3 x^{2}), at x = 0

\int 6 sec^{6} (x) d x

h (x) = (\frac{2}{x} - 3) (x^{2} + 7)

What is the integral of 1/((x-6)sqrt(x^2-12x)) ?
The integral of 1/((x-6)sqrt(x^2-12x)) is 1/6 arcsec(1/6 x-1)+C