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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

- - x^{2} + 4 x + C

Solution steps

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

Expand

\frac{x - 2}{4 x - x ^{2}} : \frac{x}{4 x - x ^{2}} - \frac{2}{4 x - x ^{2}}

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

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

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

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

Simplify

= - - x^{2} + 4 x

Add a constant to the solution

= - - x^{2} + 4 x + C

\frac{d}{d x} (\frac{x - 5}{x + 5})

s l o p e (4, - 2), (- 5, 4)

\frac{d}{d x} (- (\frac{x ^{3} + x - 1}{2}))

d er i v a t i v e f (x) = \frac{7}{x} - x

\frac{d}{d x} (1 + 2 x)

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