1. ∫(x – 1)⁴ dx 2. ∫√2x dx 3. ∫x(x² + 1)⁴ dx

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1. ∫(x – 1)⁴ dx
2. ∫√2x dx
3. ∫x(x² + 1)⁴ dx

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Jawaban Terkonfirmasi

Integral
1. ∫(x-1)⁴ dx
u = x – 1 –> du/dx = 1 –> dx = du
∫(x-1)⁴ dx = ∫ u⁴ du = ¹/₅ u⁵ + c
∫(x-1)⁴ dx = ¹/₅ (x-1)⁵ + c

2. ∫√(2x) dx
u = 2x –> du = 2 dx –> dx = 1/2 du
∫√(2x) dx = ∫ 1/2 √u du
= 1/2 ∫ u¹/² du = 1/2 (2/3)(u)³/² + c = 1/3 U³/² + c = 1/3 u √ u + c
∫√(2x) dx = 1/3 (2x) √(2x) + c
∫√(2x) dx = ²/₃ x √(2x) + c

3. ∫x(x² +1)⁴ dx
u = x² + 1 –> du = 2x dx —> x dx = 1/2 du
∫ x(x+1)⁴ dx = ∫ 1/2 u⁴ du = 1/2 (1/5)(u)⁵ + c =1/10 u⁵ + c
∫ x(x+1)⁴ dx = ¹/₁₀ (x² +1)⁵ + c