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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 2 (sin (v) + 1) d v

= \int 2 (sin (v) + 1) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int sin (v) + 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int sin (v) d v + \int 1 d v)

\int sin (v) d v = - cos (v)

\int 1 d v = v

= 2 (- cos (v) + v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int 2 x (x^{2} + 1)^{8} d x

(g (x))^{3}

x \to 0 lim (x^{3} e^{- x^{2}})

\int_{\frac{1}{2}}^{5} \frac{1}{( 2 x - 1 ) ^{\frac{1}{3}}} d x

\int 12 sin^{2} (x co) s^{2} x d x

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