What is the integral of ((x-3))/(sqrt(4x-x^2)) ?

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

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integral of ((x-3))/(sqrt(4x-x^2))

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

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

Solution steps

\int \frac{x - 3}{4 x - x ^{2}} d x

Complete the square

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

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

Apply u-substitution

= \int \frac{u - 1}{- u ^{2} + 4} d u

Expand

\frac{u - 1}{- u ^{2} + 4} : \frac{u}{- u ^{2} + 4} - \frac{1}{- u ^{2} + 4}

= \int \frac{u}{- u ^{2} + 4} - \frac{1}{- u ^{2} + 4} d u

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{u}{- u ^{2} + 4} d u - \int \frac{1}{- u ^{2} + 4} d u

\int \frac{u}{- u ^{2} + 4} d u = - - u^{2} + 4

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

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

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

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