What is the integral of (2x)/(sqrt(6x-x^2)) ?

The integral of (2x)/(sqrt(6x-x^2)) is 6(-1/3 sqrt(-x^2+6x)+arcsin(1/3 x-1))+C

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integral of (2x)/(sqrt(6x-x^2))

\int \frac{2 x}{6 x - x ^{2}} d x

6 (- \frac{1}{3} - x^{2} + 6 x + arcsin (\frac{1}{3} x - 1)) + C

Solution steps

\int \frac{2 x}{6 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}{6 x - x ^{2}} d x

Complete the square

6 x - x^{2} : - (x - 3)^{2} + 9

= 2 \cdot \int \frac{x}{- ( x - 3 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{u + 3}{- u ^{2} + 9} d u

= 2 \cdot \int \frac{u + 3}{- u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int 3 (sin (v) + 1) d v

= 2 \cdot \int 3 (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 3 \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 3 (\int sin (v) d v + \int 1 d v)

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

\int 1 d v = v

= 2 \cdot 3 (- cos (v) + v)

Substitute back

= 2 \cdot 3 (- cos (arcsin (\frac{1}{3} (x - 3))) + arcsin (\frac{1}{3} (x - 3)))

Simplify

2 \cdot 3 (- cos (arcsin (\frac{1}{3} (x - 3))) + arcsin (\frac{1}{3} (x - 3))) : 6 (- \frac{1}{3} - x^{2} + 6 x + arcsin (\frac{1}{3} x - 1))

= 6 (- \frac{1}{3} - x^{2} + 6 x + arcsin (\frac{1}{3} x - 1))

Add a constant to the solution

= 6 (- \frac{1}{3} - x^{2} + 6 x + arcsin (\frac{1}{3} x - 1)) + C

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What is the integral of (2x)/(sqrt(6x-x^2)) ?
The integral of (2x)/(sqrt(6x-x^2)) is 6(-1/3 sqrt(-x^2+6x)+arcsin(1/3 x-1))+C