definite integral 3 to 6 square root of (x+square root of 12x-36) plus square root of (x-square root of 12x-36) dx = ?

Dear

I= ₃∫⁶ {√[x+√(12x-36)]  + √[x-√(12x-36)] }dx

or

I = ₃∫⁶ {√[9-x+√(72-12x)]  + √[9-x-√(72-12x)] }dx

  =₃∫⁶ {√[3 +(6-x)+2√3√(6-x)]  + √[3+(6-x)-2√3√(6-x)] }dx

  =₃∫⁶ {|√3+√(6-x)|  + |√3 - √(6-x)| }dx

  =₃∫⁶ {√3+√(6-x)  + √3 - √(6-x) }dx

  = ₃∫⁶ 2√3dx

  =6√3

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