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

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

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

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

- - x^{2} + 6 x + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{u}{- u ^{2} + 9} d u

= \int \frac{u}{- u ^{2} + 9} d u

Apply u-substitution

: \int - \frac{1}{2 v} d v

= \int - \frac{1}{2 v} d v

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{v} d v

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= - \frac{1}{2} \cdot 2 v^{\frac{1}{2}}

Substitute back

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

Simplify

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

= - - x^{2} + 6 x

Add a constant to the solution

= - - x^{2} + 6 x + C

\frac{d}{d x} (ln (cos (x)))

\int \frac{tan ^{2} ( x )}{sec ^{5} ( x )} d x

\int \frac{w}{16 - w ^{2}} d w

\frac{d}{d x} (\frac{x}{x ^{2} + 7})

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