Solutions — Integration Rules and Composite Functions

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1. Show Solution
   ∫ (2x + 3)⁵ dx with a = 2, n = 5:
   = (2x+3)⁶ / [2 × 6] + c
   = (2x+3)⁶/12 + c
   
   Check: d/dx[(2x+3)⁶/12] = (1/12) × 6(2x+3)⁵ × 2 = (2x+3)⁵ ✓
2. Show Solution
   ∫ e^4x−1 dx with a = 4:
   = (1/4) e^4x−1 + c
   = (1/4)e^4x−1 + c
   
   Check: d/dx[(1/4)e^4x−1] = (1/4) × 4 e^4x−1 = e^4x−1 ✓
3. Show Solution
   ∫ cos(3x + π) dx with a = 3:
   = (1/3) sin(3x + π) + c
   = (1/3)sin(3x + π) + c
   
   Check: d/dx[(1/3)sin(3x+π)] = (1/3) × 3 cos(3x+π) = cos(3x+π) ✓
4. Show Solution
   ∫ 1/(5x − 2) dx with a = 5:
   = (1/5) ln|5x − 2| + c
   = (1/5)ln|5x − 2| + c
5. Show Solution
   ∫ [sin(2x) − e^3x] dx
   
   ∫ sin(2x) dx = −(1/2)cos(2x)   (a = 2)
   ∫ e^3x dx = (1/3)e^3x   (a = 3)
   
   Answer: −(1/2)cos(2x) − (1/3)e^3x + c
6. Show Solution
   ∫ (1 − 4x)³ dx with a = −4, n = 3:
   = (1−4x)⁴ / [(−4) × 4] + c
   = (1−4x)⁴ / (−16) + c
   = −(1−4x)⁴/16 + c
   
   Check: d/dx[−(1−4x)⁴/16]
   = −(1/16) × 4(1−4x)³ × (−4)
   = −(1/16) × (−16)(1−4x)³ = (1−4x)³ ✓
7. Show Solution
   ∫ 3/(2x + 7) dx = 3 ∫ 1/(2x+7) dx
   = 3 × (1/2) ln|2x+7| + c
   = (3/2)ln|2x + 7| + c
8. Show Solution
   ∫ 4cos(x/2) dx: a = 1/2, so divide by 1/2 (multiply by 2)
   = 4 × (1/(1/2)) sin(x/2) = 4 × 2 sin(x/2) = 8 sin(x/2)
   
   ∫ 2sin(3x) dx: a = 3
   = 2 × (−1/3) cos(3x) = −(2/3)cos(3x)
   
   Answer: 8 sin(x/2) − (2/3)cos(3x) + c
   
   Check: d/dx[8 sin(x/2)] = 8 × (1/2) cos(x/2) = 4 cos(x/2) ✓
   d/dx[−(2/3)cos(3x)] = −(2/3) × (−3) sin(3x) = 2 sin(3x) ✓
9. Show Solution
   Step 1 — Integrate each term:
   ∫ (2x−1)² dx = (2x−1)³/[2×3] = (2x−1)³/6
   ∫ −3e^−x/2 dx = −3 × (1/(−1/2)) e^−x/2 = −3 × (−2) e^−x/2 = 6e^−x/2
   So y = (2x−1)³/6 + 6e^−x/2 + c
   
   Step 2 — Apply y(1) = 0:
   (2×1−1)³/6 + 6e^−1/2 + c = 0
   1/6 + 6/√e + c = 0
   c = −1/6 − 6e^−1/2
   
   y = (2x−1)³/6 + 6e^−x/2 − 1/6 − 6e^−1/2
10. Show Solution
    Step 1 — Integrate:
    f(x) = ∫[sin(2x) + 1/(3x+1)] dx
    = −(1/2)cos(2x) + (1/3)ln(3x+1) + c
    (No absolute value needed since 3x + 1 > 0 for x > −1/3.)
    
    Step 2 — Apply f(0) = 2:
    −(1/2)cos(0) + (1/3)ln(1) + c = 2
    −1/2 + 0 + c = 2
    c = 5/2
    
    Step 3 — Find f(π/4):
    f(π/4) = −(1/2)cos(π/2) + (1/3)ln(3π/4 + 1) + 5/2
    = 0 + (1/3)ln(1 + 3π/4) + 5/2
    = (1/3)ln(1 + 3π/4) + 5/2