What is the integral of 3xsqrt(8x-x^2) ?

The integral of 3xsqrt(8x-x^2) is -(-x^2+8x)^{3/2}+96arcsin(1/4 x-1)+48sin(2arcsin(1/4 x-1))+C

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integral of 3xsqrt(8x-x^2)

\int 3 x 8 x - x^{2} d x

- (- x^{2} + 8 x)^{\frac{3}{2}} + 96 arcsin (\frac{1}{4} x - 1) + 48 sin (2 arcsin (\frac{1}{4} x - 1)) + C

Solution steps

\int 3 x 8 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 8 x - x^{2} d x

Complete the square

8 x - x^{2} : - (x - 4)^{2} + 16

= 3 \cdot \int x - (x - 4)^{2} + 16 d x

Apply u-substitution

= 3 \cdot \int (u + 4) - u^{2} + 16 d u

Expand

(u + 4) - u^{2} + 16 : u - u^{2} + 16 + 4 - u^{2} + 16

= 3 \cdot \int u - u^{2} + 16 + 4 - u^{2} + 16 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} + 16 d u + \int 4 - u^{2} + 16 d u)

\int u - u^{2} + 16 d u = - \frac{1}{3} (- u^{2} + 16)^{\frac{3}{2}}

\int 4 - u^{2} + 16 d u = 32 (arcsin (\frac{1}{4} u) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} u)))

= 3 (- \frac{1}{3} (- u^{2} + 16)^{\frac{3}{2}} + 32 (arcsin (\frac{1}{4} u) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} u))))

Substitute back

u = x - 4

= 3 (- \frac{1}{3} (- (x - 4)^{2} + 16)^{\frac{3}{2}} + 32 (arcsin (\frac{1}{4} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 4)))))

Simplify

3 (- \frac{1}{3} (- (x - 4)^{2} + 16)^{\frac{3}{2}} + 32 (arcsin (\frac{1}{4} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 4))))) : - (- x^{2} + 8 x)^{\frac{3}{2}} + 96 arcsin (\frac{1}{4} x - 1) + 48 sin (2 arcsin (\frac{1}{4} x - 1))

= - (- x^{2} + 8 x)^{\frac{3}{2}} + 96 arcsin (\frac{1}{4} x - 1) + 48 sin (2 arcsin (\frac{1}{4} x - 1))

Add a constant to the solution

= - (- x^{2} + 8 x)^{\frac{3}{2}} + 96 arcsin (\frac{1}{4} x - 1) + 48 sin (2 arcsin (\frac{1}{4} x - 1)) + C

What is the integral of 3xsqrt(8x-x^2) ?
The integral of 3xsqrt(8x-x^2) is -(-x^2+8x)^{3/2}+96arcsin(1/4 x-1)+48sin(2arcsin(1/4 x-1))+C