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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

Expand

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

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

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

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

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

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

\int_{e}^{e^{2}} \frac{1}{x ln ( x )} d x

\frac{d}{d x} (\frac{40}{1 + 79 e ^{- x}})

\frac{d}{d x} (\frac{x}{x - 7})

\int \frac{2 x - 1}{4 x ^{2} - 1} d x

\frac{\partial}{\partial y} (5 x y^{2})

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