₀^2 3 x+1 x^2+4 d x=

Step-by-step solution

The integral I = ₀^2 3x+1 x^2+4 dx separates into two components: I = ₀^2 3x x^2+4 dx + ₀^2 1 x^2+4 dx For the first term, substitute u = x^2 + 4 so that du = 2x dx and x dx = 1 2 du : When x = 0 , u = 4 When x = 2 , u = 8 This yields 3 2 ₄^8 1 u du = 3 2 |u | |₄^8 = 3 2 ( 8 - 4) = 3 2 2 = (2 2 ) The second term follows the standard form: ₀^2 1 x^2+4 dx = 1 2 ( x 2 ) |₀^2 = 1 2 ( 4 - 0 ) = 8 The complete integral evaluates to (2 2 ) + 8 , matching option C .