a. Using u-substitution, let u = 3x+6. Then du/dx = 3 and dx = du/3. Substituting, we get:
S dx x 3x+6 = S (u-6)/3 du = (1/3) S (u-6) du
= (1/3) [(u²/2) - 6u] + C
= (1/6) (3x+6)² - 2(3x+6) + C
= (1/6) (9x² + 36x + 36) - 6x - 12 + C
= (3/2) x² + 3x - 2 + C
b. Expanding (2x²+8x+3)², we get:
S (2x²+8x+3)² dx = S (4x⁴ + 32x³ + 82x² + 48x + 9) dx
= (4/5) x⁵ + (8/3) x⁴ + (82/3) x³ + 24x² + 9x + C
c. Using u-substitution, let u = -x². Then du/dx = -2x and dx = -du/(2x). Substituting, we get:
S 5xe-x² dx = -5 S e^u du/(2x) = (-5/2) S e^u du/u
= (-5/2) ln|-x²| + C
= (-5/2) ln(x²) + C
= -5 ln|x| + C
d. Using the power rule, we get:
S (y^5+1) dx = (1/6) y^6 + y + C