What is the integral of 3xsqrt(4x-x^2) ?

The integral of 3xsqrt(4x-x^2) is -(-x^2+4x)^{3/2}+12arcsin(1/2 x-1)+6sin(2arcsin(1/2 x-1))+C

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integral of 3xsqrt(4x-x^2)

\int 3 x 4 x - x^{2} d x

- (- x^{2} + 4 x)^{\frac{3}{2}} + 12 arcsin (\frac{1}{2} x - 1) + 6 sin (2 arcsin (\frac{1}{2} x - 1)) + C

Solution steps

\int 3 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

= 3 \cdot \int x 4 x - x^{2} d x

Complete the square

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

= 3 \cdot \int x - (x - 2)^{2} + 4 d x

Apply u-substitution

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

= 3 \cdot \int (u + 2) - u^{2} + 4 d u

Expand

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

= 3 \cdot \int u - u^{2} + 4 + 2 - 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

= 3 (\int u - u^{2} + 4 d u + \int 2 - u^{2} + 4 d u)

\int u - u^{2} + 4 d u = - \frac{1}{3} (- u^{2} + 4)^{\frac{3}{2}}

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

= 3 (- \frac{1}{3} (- u^{2} + 4)^{\frac{3}{2}} + 4 (arcsin (\frac{1}{2} u) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} u))))

Substitute back

u = x - 2

= 3 (- \frac{1}{3} (- (x - 2)^{2} + 4)^{\frac{3}{2}} + 4 (arcsin (\frac{1}{2} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 2)))))

Simplify

3 (- \frac{1}{3} (- (x - 2)^{2} + 4)^{\frac{3}{2}} + 4 (arcsin (\frac{1}{2} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 2))))) : - (- x^{2} + 4 x)^{\frac{3}{2}} + 12 arcsin (\frac{1}{2} x - 1) + 6 sin (2 arcsin (\frac{1}{2} x - 1))

= - (- x^{2} + 4 x)^{\frac{3}{2}} + 12 arcsin (\frac{1}{2} x - 1) + 6 sin (2 arcsin (\frac{1}{2} x - 1))

Add a constant to the solution

= - (- x^{2} + 4 x)^{\frac{3}{2}} + 12 arcsin (\frac{1}{2} x - 1) + 6 sin (2 arcsin (\frac{1}{2} x - 1)) + C

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What is the integral of 3xsqrt(4x-x^2) ?
The integral of 3xsqrt(4x-x^2) is -(-x^2+4x)^{3/2}+12arcsin(1/2 x-1)+6sin(2arcsin(1/2 x-1))+C