Integration

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Integration is the inverse operation of differentiation. If f(x) and g(x) are two functions such that d/dx[f(x)] = g(x), then f(x) is called an integral of g(x) with respect to x. It is denoted by ∫g(

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1. ∫k·f(x)dx = k∫f(x)dx (constant multiple rule) 2. ∫[f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx (sum rule) 3. ∫[f(x) - g(x)]dx = ∫f(x)dx - ∫g(x)dx (difference rule) These rules allow us to integrate linear c

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∫x^n dx = x^(n+1)/(n+1) + c, where n ≠ -1 Condition: This formula is valid for all real numbers n except n = -1. For n = -1: ∫(1/x)dx = log|x| + c Example: ∫x³dx = x⁴/4 + c

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Step 1: Apply the sum/difference rule ∫(3x² - 5x + 2)dx = ∫3x²dx - ∫5x dx + ∫2 dx Step 2: Apply constant multiple rule = 3∫x²dx - 5∫x dx + 2∫dx Step 3: Use standard formulas = 3(x³/3) - 5(x²/2) + 2x

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The substitution method (Method of Change of Variable) is used to simplify complex integrals by changing the variable. We substitute x = φ(t) where φ(t) is a differentiable function. Theorem: ∫f(x)dx

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Step 1: Let u = 2x + 1, then du = 2dx, so dx = du/2 Step 2: Substitute ∫(2x + 1)⁵dx = ∫u⁵ · (du/2) = (1/2)∫u⁵du Step 3: Integrate = (1/2) · u⁶/6 + c = u⁶/12 + c Step 4: Back substitute = (2x + 1)⁶/

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Corollary: ∫f(x)ⁿ·f'(x)dx = [f(x)]^(n+1)/(n+1) + c, where n ≠ -1 Example: ∫x²·√(x³ + 1)dx Here f(x) = x³ + 1, so f'(x) = 3x² Rewrite as: ∫(x³ + 1)^(1/2) · 3x²dx · (1/3) = (1/3) ∫(x³ + 1)^(1/2) · 3x²d

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Formula: ∫f'(x)/f(x)dx = log|f(x)| + c Example: ∫(2x)/(x² + 1)dx Here f(x) = x² + 1, so f'(x) = 2x Therefore: ∫(2x)/(x² + 1)dx = log|x² + 1| + c = log(x² + 1) + c (Since x² + 1 > 0 for all real x) A