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

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

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

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

Simplify

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

Add a constant to the solution

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

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

f (θ) = 5 θ

\int (14 x + 15 x) d x

f (x) = 7 x - 4 x^{2}, (- 1, - 11)

f (x) = \frac{5}{x - 7}, (8, 5)

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