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)+4arcsin(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+2))/(sqrt(4x-x^2))

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{60} w h

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

\frac{d}{d x} (\frac{64}{x})

f (x) = 8 x^{2} + 5 x, at x = 3

\frac{d}{d x} (cos^{3 (\frac{x ^{2}}{x - 1})} (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)+4arcsin(1/2 x-1)+C