Using the most appropriate method, evaluate the following integrals: (a) ∫x(3x2 + 2)2 dx (b) ∫ (3x2 - 1)/(x3 - x + 4)3 dx (c) ∫_-11 (x-1)(x+3)dx

Here are the solutions to the integration problems: Question 1 (a) a) To evaluate x(3x^2 + 2)^2 dx, we use u-substitution. Step 1: Let u = 3x^2 + 2. Step 2: Differentiate u with respect to x to find du: du = (d)/(dx)(3x^2 + 2) dx = 6x \, dx Step 3: Rearrange to find x \, dx: x \, dx = (1)/(6) du Step 4: Substitute u and x \, dx into the integral: u^2 ((1)/(6)) du = (1)/(6) u^2 du Step 5: Integrate with respect to u: (1)/(6) (u^2+12+1) + C = (1)/(6) ((u^3)/(3)) + C = (u^3)/(18) + C Step 6: Substitute back u = 3x^2 + 2: ((3x^2 + 2)^3)/(18) + C b) To evaluate (3x^2 - 1)/((x^3 - x + 4)^3) dx, we use u-substitution. Step 1: Let u = x^3 - x + 4. Step 2: Differentiate u with respect to x to find du: du = (d)/(dx)(x^3 - x + 4) dx = (3x^2 - 1) \, dx Step 3: Substitute u and du into the integral: (1)/(u^3) du = u^-3 du Step 4: Integrate with respect to u: u^-3+1-3+1 + C = u^-2-2 + C = -(1)/(2u^2) + C Step 5: Substitute back u = x^3 - x + 4: -(1)/(2(x^3 - x + 4)^2) + C c) To evaluate _-1^1 (x - 1)(x + 3) dx, we first expand the integrand. Step 1: Expand the product (x - 1)(x + 3): (x - 1)(x + 3) = x^2 + 3x - x - 3 = x^2 + 2x - 3 Step 2: Integrate the polynomial: (x^2 + 2x - 3) dx = (x^3)/(3) + (2x^2)/(2) - 3x = (x^3)/(3) + x^2 - 3x Step 3: Evaluate the definite integral using the limits from -1 to 1: [ (x^3)/(3) + x^2 - 3x ]_-1^1 = ( ((1)^3)/(3) + (1)^2 - 3(1) ) - ( ((-1)^3)/(3) + (-1)^2 - 3(-1) ) Step 4: Calculate the values at the upper and lower limits: = ( (1)/(3) + 1 - 3 ) - ( -(1)/(3) + 1 + 3 ) = ( (1)/(3) - 2 ) - ( -(1)/(3) + 4 ) = ( (1)/(3) - (6)/(3) ) - ( -(1)/(3) + (12)/(3) ) = -(5)/(3) - (11)/(3) Step 5: Subtract the values: = -(16)/(3) -(16)/(3) Question 1 (b) a) To integrate ^5 x using Pythagoras' theorem of odd powers: Step 1: Rewrite ^5 x as ^4 x x: ^5 x \, dx = ^4 x x \, dx Step 2: Use the identity ^2 x = 1 - ^2 x: = (^2 x)^2 x \, dx = (1 - ^2 x)^2 x \, dx Step 3: Let u = x. Then du = - x \, dx, so x \, dx = -du. = (1 - u^2)^2 (-du) = - (1 - 2u^2 + u^4) du Step 4: Integrate the polynomial with respect to u: = -( u - (2u^3)/(3) + (u^5)/(5) ) + C Step 5: Substitute back u = x: x + (2)/(3)^3 x - (1)/(5)^5 x + C b) To integrate ^3 x using Pythagoras' theorem of odd powers: Step 1: Rewrite ^3 x as ^2 x x: ^3 x \, dx = ^2 x x \, dx Step 2: Use the identity ^2 x = 1 - ^2 x: = (1 - ^2 x) x \, dx Step 3: Let u = x. Then du = x \, dx. = (1 - u^2) du Step 4: Integrate the polynomial with respect to u: = u - (u^3)/(3) + C Step 5: Substitute back u = x: x - (1)/(3)^3 x + C What's next? Send 'em! 📸