What is the integral of 1/(sqrt((x-9)(3-x))) ?

The integral of 1/(sqrt((x-9)(3-x))) is arcsin(1/3 (x-6))+C

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integral of 1/(sqrt((x-9)(3-x)))

\int \frac{1}{( x - 9 ) ( 3 - x )} d x

arcsin (\frac{1}{3} (x - 6)) + C

Solution steps

\int \frac{1}{( x - 9 ) ( 3 - x )} d x

\frac{1}{( x - 9 ) ( 3 - x )} = \frac{1 - ( x - 9 ) ( 3 - x )}{( x - 9 ) ( 3 - x )} + 1

= \int \frac{1 - ( x - 9 ) ( 3 - x )}{( x - 9 ) ( 3 - x )} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1 - ( x - 9 ) ( 3 - x )}{( x - 9 ) ( 3 - x )} d x + \int 1 d x

\int \frac{1 - ( x - 9 ) ( 3 - x )}{( x - 9 ) ( 3 - x )} d x = arcsin (\frac{1}{3} (x - 6)) - x

\int 1 d x = x

= arcsin (\frac{1}{3} (x - 6)) - x + x

Simplify

= arcsin (\frac{1}{3} (x - 6))

Add a constant to the solution

= arcsin (\frac{1}{3} (x - 6)) + C

x y^{^{''}} + 2 y^{^{'}} = 24 x^{2}

\int - 5 x d x

y = (3 x^{2} + 2) (4 x - 1)

\frac{d}{d x} (\frac{e ^{x}}{x ^{2}})

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

What is the integral of 1/(sqrt((x-9)(3-x))) ?
The integral of 1/(sqrt((x-9)(3-x))) is arcsin(1/3 (x-6))+C