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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{x ^{2}}{4 x - x ^{2}} d x = 6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x

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

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

= 6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x - 8 (- \frac{1}{2} - x^{2} + 4 x + arcsin (\frac{1}{2} x - 1)) + 4 arcsin (\frac{1}{2} x - 1)

Simplify

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

= 2 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1))

Add a constant to the solution

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

\frac{\partial}{\partial x} (2 x y^{3} + 2 x y - 4)

f (x) = \frac{x + 2 x ^{2}}{1 + x ^{2}}

\frac{\partial}{\partial x} (sin (6 x - 7 y))

\int \frac{cos ^{3} ( 2 x )}{sin ( 2 x )} d x

\frac{π}{4}

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