What is the integral of (x-3)/(sqrt(9-(x-3)^2)) ?

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

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integral of (x-3)/(sqrt(9-(x-3)^2))

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

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

Solution steps

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

Apply u-substitution

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

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}{u} d u

Apply the Power Rule

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

Substitute back

u = 9 - (x - 3)^{2}

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

Simplify

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

= - - x^{2} + 6 x

Add a constant to the solution

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

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