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

The integral of 1/((x^2-6x)^{3/2)} is -(x-3)/(9(x^2-6x)^{1/2)}+C

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integral of 1/((x^2-6x)^{3/2)}

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

- \frac{x - 3}{9 ( x ^{2} - 6 x ) ^{\frac{1}{2}}} + C

Solution steps

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

Complete the square

x^{2} - 6 x : (x - 3)^{2} - 9

= \int \frac{1}{( ( x - 3 ) ^{2} - 9 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( u ^{2} - 9 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( u ^{2} - 9 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{9 tan ^{2} ( v )} d v

= \int \frac{sec ( v )}{9 tan ^{2} ( v )} 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 \frac{sec ( v )}{tan ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{9} \cdot \int \frac{cos ( v )}{sin ^{2} ( v )} d v

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

= \frac{1}{9} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= \frac{1}{9} (- \frac{1}{w})

Substitute back

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

Simplify

\frac{1}{9} (- \frac{1}{sin ( arcsec ( \frac{1}{3} ( x - 3 ) ) )}) : - \frac{x - 3}{9 ( x ^{2} - 6 x ) ^{\frac{1}{2}}}

= - \frac{x - 3}{9 ( x ^{2} - 6 x ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x - 3}{9 ( x ^{2} - 6 x ) ^{\frac{1}{2}}} + C

\frac{\partial}{\partial x} (x y E (x, y) (x, y)^{\frac{x}{y}})

\int - 2 cos^{2} (x) d x

f (x) = 2 sec (3 x^{2})

(6 y + 2 t - 7) d t + (8 y + 6 t - 1) d y = 0, y (- 1) = 2

\int_{0}^{t} t^{2} e^{- \frac{t}{2}} d t

What is the integral of 1/((x^2-6x)^{3/2)} ?
The integral of 1/((x^2-6x)^{3/2)} is -(x-3)/(9(x^2-6x)^{1/2)}+C