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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

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

\int 1 d v = v

= 2 \cdot 2 (- cos (v) + v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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