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

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

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

f (1) = (x^{3} + 2 x + 8)^{3}

\frac{x ^{2}}{x - 3}

\int ln (x - 20) d 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