What is the integral of (x-3)/(sqrt(8x-x^2)) ?

The integral of (x-3)/(sqrt(8x-x^2)) is -sqrt(-x^2+8x)+arcsin(1/4 (x-4))+C

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integral of (x-3)/(sqrt(8x-x^2))

\int \frac{x - 3}{8 x - x ^{2}} d x

- - x^{2} + 8 x + arcsin (\frac{1}{4} (x - 4)) + C

Solution steps

\int \frac{x - 3}{8 x - x ^{2}} d x

Complete the square

8 x - x^{2} : - (x - 4)^{2} + 16

= \int \frac{x - 3}{- ( x - 4 ) ^{2} + 16} d x

Apply u-substitution

= \int \frac{u}{- u ^{2} + 2 u + 15} d u

Complete the square

- u^{2} + 2 u + 15 : - (u - 1)^{2} + 16

= \int \frac{u}{- ( u - 1 ) ^{2} + 16} d u

Apply u-substitution

= \int \frac{v + 1}{- v ^{2} + 16} d v

Expand

\frac{v + 1}{- v ^{2} + 16} : \frac{v}{- v ^{2} + 16} + \frac{1}{- v ^{2} + 16}

= \int \frac{v}{- v ^{2} + 16} + \frac{1}{- v ^{2} + 16} 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

= \int \frac{v}{- v ^{2} + 16} d v + \int \frac{1}{- v ^{2} + 16} d v

\int \frac{v}{- v ^{2} + 16} d v = - - v^{2} + 16

\int \frac{1}{- v ^{2} + 16} d v = arcsin (\frac{1}{4} v)

= - - v^{2} + 16 + arcsin (\frac{1}{4} v)

Substitute back

= - - (x - 3 - 1)^{2} + 16 + arcsin (\frac{1}{4} (x - 3 - 1))

Simplify

- - (x - 3 - 1)^{2} + 16 + arcsin (\frac{1}{4} (x - 3 - 1)) : - - x^{2} + 8 x + arcsin (\frac{1}{4} (x - 4))

= - - x^{2} + 8 x + arcsin (\frac{1}{4} (x - 4))

Add a constant to the solution

= - - x^{2} + 8 x + arcsin (\frac{1}{4} (x - 4)) + C

What is the integral of (x-3)/(sqrt(8x-x^2)) ?
The integral of (x-3)/(sqrt(8x-x^2)) is -sqrt(-x^2+8x)+arcsin(1/4 (x-4))+C