What is the integral of 1/((x-9)sqrt(x^2-18x)) ?

The integral of 1/((x-9)sqrt(x^2-18x)) is 1/9 arcsec(1/9 x-1)+C

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integral of 1/((x-9)sqrt(x^2-18x))

\int \frac{1}{( x - 9 ) x ^{2} - 18 x} d x

\frac{1}{9} arcsec (\frac{1}{9} x - 1) + C

Solution steps

\int \frac{1}{( x - 9 ) x ^{2} - 18 x} d x

Apply u-substitution

= \int \frac{1}{u u ^{2} - 81} d u

Apply Trigonometric Substitution

= \int \frac{1}{9} d v

Take the constant out:

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

= \frac{1}{9} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{9} \cdot 1 \cdot v

Substitute back

= \frac{1}{9} \cdot 1 \cdot arcsec (\frac{1}{9} (x - 9))

Simplify

\frac{1}{9} \cdot 1 \cdot arcsec (\frac{1}{9} (x - 9)) : \frac{1}{9} arcsec (\frac{1}{9} x - 1)

= \frac{1}{9} arcsec (\frac{1}{9} x - 1)

Add a constant to the solution

= \frac{1}{9} arcsec (\frac{1}{9} x - 1) + C

e x p an d \frac{2}{( - x ^{3} + 9 x ) ^{4}}

\int \frac{x}{1 + x} d x

y^{^{'}} = 1 + x - y - x y, y (0) = - 1

x \to π lim (- 4 sin (\frac{x}{2}))

(1 - x^{2}) y^{^{'}} = 1

What is the integral of 1/((x-9)sqrt(x^2-18x)) ?
The integral of 1/((x-9)sqrt(x^2-18x)) is 1/9 arcsec(1/9 x-1)+C