ĂSAE 6x sin 3xdx= 2x cos 3x+2//3sin 3x+c -2x cos 3x+2//3sin 3x+C -2x cos 3x-2//3sin 3x+C -2//3x cos 3x-2//3sin 3x+C

B

To solve the integral of \(6x \sin(3x) dx\), we integrate the function with respect to . The result is \( \frac{2}{3} \sin(3x) - 2x \cos(3x) + C \), where is the constant of integration. This matches option B: \(-2x \cos(3x) + \frac{2}{3} \sin(3x) + C\).