Indefinite Integration

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Indefinite integration is the reverse process of differentiation. If f(x) is a given function, we find g(x) such that g'(x) = f(x). The integral of f(x) with respect to x is g(x), written as ∫f(x)dx =

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Power Rule: ∫x^n dx = x^(n+1)/(n+1) + c, where n ≠ -1 Step-by-step application: 1. Add 1 to the power: n becomes (n+1) 2. Divide by the new power: (n+1) 3. Add constant c Example: ∫x⁵ dx = x⁶/6 + c

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Step-by-step solution: ∫(3x² + 2x - 5) dx = ∫3x² dx + ∫2x dx - ∫5 dx = 3∫x² dx + 2∫x dx - 5∫1 dx = 3(x³/3) + 2(x²/2) - 5x + c = x³ + x² - 5x + c Key concept: Use linearity property - integral of sum

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Standard Trigonometric Integrals: • ∫cos x dx = sin x + c • ∫sin x dx = -cos x + c • ∫sec² x dx = tan x + c • ∫cosec² x dx = -cot x + c • ∫sec x tan x dx = sec x + c • ∫cosec x cot x dx = -cosec x + c

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Step-by-step solution: ∫tan x dx = ∫(sin x/cos x) dx Method: Rewrite as ratio and use substitution = ∫(sin x/cos x) dx = -∫(-sin x/cos x) dx Let cos x = t, then -sin x dx = dt = -∫(1/t) dt = -log|t|

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Substitution Theorem: If x = φ(t), then ∫f(x)dx = ∫f[φ(t)]φ'(t)dt Example: ∫3x²sin(x³)dx Step 1: Let x³ = t Step 2: Differentiate: 3x²dx = dt Step 3: Substitute: ∫sin(t)dt = -cos(t) + c Step 4: Back-

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Step-by-step solution: ∫(2x+3)/(x²+3x+1) dx Step 1: Notice that d/dx(x²+3x+1) = 2x+3 Step 2: Let u = x²+3x+1, then du = (2x+3)dx Step 3: ∫(2x+3)/(x²+3x+1) dx = ∫(1/u) du Step 4: = log|u| + c Step 5:

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Standard Forms: 1. ∫1/√(a²-x²) dx = sin⁻¹(x/a) + c 2. ∫1/√(a²+x²) dx = log|x + √(a²+x²)| + c 3. ∫1/√(x²-a²) dx = log|x + √(x²-a²)| + c 4. ∫1/(a²+x²) dx = (1/a)tan⁻¹(x/a) + c 5. ∫1/(a²-x²) dx = (1/2a)l