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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u + 3}{9 - u ^{2}} d u

= \int \frac{u + 3}{9 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 3 (sin (v) + 1) d v

= \int 3 (sin (v) + 1) d v

Take the constant out:

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

= 3 \cdot \int sin (v) + 1 d v

Apply the Sum Rule:

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

= 3 (\int sin (v) d v + \int 1 d v)

\int sin (v) d v = - cos (v)

\int 1 d v = v

= 3 (- cos (v) + v)

Substitute back

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

Simplify

3 (- cos (arcsin (\frac{1}{3} (x - 3))) + arcsin (\frac{1}{3} (x - 3))) : - - x^{2} + 6 x + 3 arcsin (\frac{1}{3} x - 1)

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

Add a constant to the solution

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

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