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Maths

11 Dec 2025

62 pages

Understanding Integral Calculus Basics

Elysa Mae Tanteo @elysamaetanteo

Ever wondered how to work backwards from derivatives? Integral calculusis basically differentiation in reverse - it's your... Show more

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Introduction to Integral Calculus

Think of anti-differentiation as detective work - you're given clues (the derivative) and need to find the original function. The process is written as ∫f(x)dx = F(x) + C, where the curvy ∫ symbol means "integrate this."

Here's the brilliant bit if you differentiate y = x³ + 12, y = x³ - 300, or just y = x³, you always get dy/dx = 3x². This means when you integrate 3x² backwards, you could get any of these functions.

That's why we add C (the constant of integration) - it represents all those possible constants that disappear during differentiation. When you see ∫3x²dx = x³ + C, that C could be 12, -300, or any number at all.

Key insight Integration is the inverse of differentiation - they undo each other perfectly!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Essential Integration Formulas

These basic integration formulas are your mathematical toolkit - memorise them and you'll tackle most problems with confidence. The simplest ones involve constants and powers ∫dx = x + C and ∫adx = ax + C.

The power rule is your best friend ∫xⁿdx = xⁿ⁺¹/ $n+1$ + C. Just add 1 to the power and divide by the new power. For functions like sine and cosine, remember that ∫cos x dx = sin x + C, whilst ∫sin x dx = -cos x + C (note the minus sign!).

Don't forget the special cases ∫ $1/x$ dx = ln|x| + C for logarithmic functions, and ∫eˣdx = eˣ + C for exponentials. The general power rule ∫uⁿdu = uⁿ⁺¹/ $n+1$ + C works when u represents any function.

Pro tip The power rule fails when n = -1, which is why we have the separate logarithmic formula!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Working with Power Rules

Let's see these formulas in action with real examples that show you exactly how the power rule works in practice. For ∫x⁷dx, you get x⁸/8 + C by adding 1 to the power (making it 8) and dividing by 8.

Negative powers work the same way ∫x⁻⁵dx becomes x⁻⁴/(-4) + C. Fractional powers need a bit more care - remember that ∫x^(1/2)dx = x^(3/2)/(3/2) + C, which simplifies to $2x^(3/2)$ /3 + C.

The trickiest bit is dealing with expressions like ∫ $1/x⁵$ dx. First, rewrite this as ∫x⁻⁵dx, then apply the power rule normally. Always convert roots and fractions to power notation first - it makes everything cleaner.

Remember Always convert fractions and roots to power notation before integrating - it prevents silly mistakes!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Constant Multiples and Sum Rules

Constants make integration easier, not harder - you can pull them outside the integral sign. So ∫3x²dx becomes 3∫x²dx = 3 $x³/3$ + C = x³ + C. The constant just tags along for the ride.

The sum and difference rule lets you split complex expressions ∫ $3x² + 9x - 5$ dx becomes ∫3x²dx + ∫9x dx - ∫5dx. Each piece integrates separately, giving you x³ + (9x²)/2 - 5x + C.

This approach works brilliantly for polynomials like ∫ $4x³ - 5x² + 4x$ dx. Break it down term by term 4∫x³dx - 5∫x²dx + 4∫x dx = x⁴ - (5x³)/3 + 2x² + C.

Study hack Always tackle one term at a time - it's much less overwhelming than trying to integrate everything at once!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Advanced Power Rules and Substitution

When you encounter expressions like ∫ $5x² + 4$ ²dx, you've got two choices expand it out using FOIL or use the general power rule. Expanding gives $25x⁴ + 40x² + 16$ , which integrates to 5x⁵ + (40x³)/3 + 16x + C.

The general power rule ∫uⁿdu = uⁿ⁺¹/ $n+1$ + C becomes crucial when dealing with composite functions. This is where substitution starts to shine - you'll identify what u represents and find its derivative du.

For expressions involving logarithms, remember that ∫x⁻¹dx = ∫ $1/x$ dx = ln|x| + C. This special case handles the power rule exception perfectly.

Game changer Learning to spot when to use substitution versus direct integration will save you hours of work!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Substitution Method in Action

The substitution method transforms tricky integrals into manageable ones by clever variable changes. For ∫ $e^(3x+4)$ ⁵ · 2e^(3x)dx, set u = e^ $3x+4$ , so du = 3e^(3x)dx, making 2e^(3x)dx = (2/3)du.

Your integral becomes (2/3)∫u⁵du = (2/3) · u⁶/6 + C = u⁶/9 + C. Substitute back to get $e^(3x+4)$ ⁶/9 + C. The key is recognising that part of your integrand matches the derivative of another part.

For trigonometric examples like ∫ $3sin x + 5$ ⁴cos x dx, use u = 3sin x + 5 and du = 3cos x dx. This transforms your integral into (1/3)∫u⁴du, which equals u⁵/15 + C.

Success tip Always check if part of your integrand looks like the derivative of another part - that's your cue to use substitution!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

More Substitution Examples

Complex polynomials become straightforward with the right substitution approach. For ∫ $4x² + 2x$ ⁷ · $12x + 2$ dx, notice that the derivative of 4x² + 2x is 12x + 2. Set u = 4x² + 2x, then du = $12x + 2$ dx.

Your integral simplifies to ∫u⁷du = u⁸/8 + C = $4x² + 2x$ ⁸/8 + C. Trigonometric substitutions work similarly - for ∫tan⁷x sec²x dx, use u = tan x since du = sec²x dx.

The pattern emerges clearly ∫sin⁸(4x)cos(4x)dx uses u = sin(4x) and du = 4cos(4x)dx. This gives you (1/4)∫u⁸du = u⁹/36 + C = (sin(4x))⁹/36 + C.

Pattern recognition Once you've done a few substitutions, you'll start spotting the derivative relationships automatically!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Logarithmic Integration Techniques

The chain rule counterpart ∫ $du/u$ = ln|u| + C handles fractions where the numerator is the derivative of the denominator. For ∫dy/ $3(y+12)$ , factor out 1/3 and set u = y + 12, so du = dy.

This becomes (1/3)∫ $du/u$ = (1/3)ln|y + 12| + C. The absolute value signs ensure we can handle negative values inside the logarithm safely.

More complex examples like ∫(2dx)/ $√x(2 + √x)$ require careful substitution. Set u = 2 + √x, then du = dx/(2√x), making dx/√x = 2du. Your integral becomes 4∫ $du/u$ = 4ln|2 + √x| + C.

Essential rule When you see a fraction with the derivative of the denominator in the numerator, think logarithmic integration!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Exponential and Trigonometric Cases

Logarithmic integration shines with exponential functions too. For ∫ $e^x dx$ / $1 + e^x$ , set u = 1 + e^x, so du = e^x dx. This transforms into ∫ $du/u$ = ln|1 + e^x| + C.

Trigonometric cases follow the same pattern. In ∫(sin 3x dx)/ $4 - cos 3x$ , use u = 4 - cos 3x, giving du = 3sin 3x dx. Since you only have sin 3x dx, multiply by 1/3 to get (-1/3)ln|4 - cos 3x| + C.

The negative sign appears because du = 3sin 3x dx, but you need sin 3x dx = (1/3)du, and the derivative of -cos 3x gives you +3sin 3x dx.

Watch out Pay careful attention to signs when working with trigonometric functions - they can flip unexpectedly!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

Advanced Logarithmic Integration

Sometimes you'll encounter quotients involving more complex trigonometric expressions. For ∫(sec²4x dx)/(5tan 4x), factor out 1/5 first, then recognise that d/dx(tan 4x) = 4sec²4x.

Set u = tan 4x, so du = 4sec²4x dx, making sec²4x dx = (1/4)du. Your integral becomes (1/5) · (1/4)∫ $du/u$ = (1/20)ln|tan 4x| + C.

The absolute value in ln|u| is crucial because tangent can be negative, and you can't take the logarithm of negative numbers. This keeps your answer mathematically sound across all domains.

Final tip Always include absolute value signs in logarithmic integration - it's a common exam mistake to forget them!

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11 Dec 2025

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62 pages

Understanding Integral Calculus Basics

Elysa Mae Tanteo

@elysamaetanteo

Ever wondered how to work backwards from derivatives? Integral calculusis basically differentiation in reverse - it's your key to finding original functions when you only know their rates of change. This essential maths skill will unlock everything from area... Show more

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

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Introduction to Integral Calculus

Here's the brilliant bit: if you differentiate y = x³ + 12, y = x³ - 300, or just y = x³, you always get dy/dx = 3x². This means when you integrate 3x² backwards, you could get any of these functions.

Key insight: Integration is the inverse of differentiation - they undo each other perfectly!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

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Essential Integration Formulas

The power rule is your best friend: ∫xⁿdx = xⁿ⁺¹/ $n+1$ + C. Just add 1 to the power and divide by the new power. For functions like sine and cosine, remember that ∫cos x dx = sin x + C, whilst ∫sin x dx = -cos x + C (note the minus sign!).

Don't forget the special cases: ∫ $1/x$ dx = ln|x| + C for logarithmic functions, and ∫eˣdx = eˣ + C for exponentials. The general power rule ∫uⁿdu = uⁿ⁺¹/ $n+1$ + C works when u represents any function.

Pro tip: The power rule fails when n = -1, which is why we have the separate logarithmic formula!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

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Working with Power Rules

Negative powers work the same way: ∫x⁻⁵dx becomes x⁻⁴/(-4) + C. Fractional powers need a bit more care - remember that ∫x^(1/2)dx = x^(3/2)/(3/2) + C, which simplifies to $2x^(3/2)$ /3 + C.

Remember: Always convert fractions and roots to power notation before integrating - it prevents silly mistakes!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

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Constant Multiples and Sum Rules

Constants make integration easier, not harder - you can pull them outside the integral sign. So ∫3x²dx becomes 3∫x²dx = 3 $x³/3$ + C = x³ + C. The constant just tags along for the ride.

The sum and difference rule lets you split complex expressions: ∫ $3x² + 9x - 5$ dx becomes ∫3x²dx + ∫9x dx - ∫5dx. Each piece integrates separately, giving you x³ + (9x²)/2 - 5x + C.

This approach works brilliantly for polynomials like ∫ $4x³ - 5x² + 4x$ dx. Break it down term by term: 4∫x³dx - 5∫x²dx + 4∫x dx = x⁴ - (5x³)/3 + 2x² + C.

Study hack: Always tackle one term at a time - it's much less overwhelming than trying to integrate everything at once!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

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Advanced Power Rules and Substitution

When you encounter expressions like ∫ $5x² + 4$ ²dx, you've got two choices: expand it out using FOIL or use the general power rule. Expanding gives $25x⁴ + 40x² + 16$ , which integrates to 5x⁵ + (40x³)/3 + 16x + C.

For expressions involving logarithms, remember that ∫x⁻¹dx = ∫ $1/x$ dx = ln|x| + C. This special case handles the power rule exception perfectly.

Game changer: Learning to spot when to use substitution versus direct integration will save you hours of work!

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Substitution Method in Action

For trigonometric examples like ∫ $3sin x + 5$ ⁴cos x dx, use u = 3sin x + 5 and du = 3cos x dx. This transforms your integral into (1/3)∫u⁴du, which equals u⁵/15 + C.

Success tip: Always check if part of your integrand looks like the derivative of another part - that's your cue to use substitution!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

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More Substitution Examples

Your integral simplifies to ∫u⁷du = u⁸/8 + C = $4x² + 2x$ ⁸/8 + C. Trigonometric substitutions work similarly - for ∫tan⁷x sec²x dx, use u = tan x since du = sec²x dx.

The pattern emerges clearly: ∫sin⁸(4x)cos(4x)dx uses u = sin(4x) and du = 4cos(4x)dx. This gives you (1/4)∫u⁸du = u⁹/36 + C = (sin(4x))⁹/36 + C.

Pattern recognition: Once you've done a few substitutions, you'll start spotting the derivative relationships automatically!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

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Logarithmic Integration Techniques

The chain rule counterpart ∫ $du/u$ = ln|u| + C handles fractions where the numerator is the derivative of the denominator. For ∫dy/ $3(y+12)$ , factor out 1/3 and set u = y + 12, so du = dy.

This becomes (1/3)∫ $du/u$ = (1/3)ln|y + 12| + C. The absolute value signs ensure we can handle negative values inside the logarithm safely.

Essential rule: When you see a fraction with the derivative of the denominator in the numerator, think logarithmic integration!

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Exponential and Trigonometric Cases

Logarithmic integration shines with exponential functions too. For ∫ $e^x dx$ / $1 + e^x$ , set u = 1 + e^x, so du = e^x dx. This transforms into ∫ $du/u$ = ln|1 + e^x| + C.

The negative sign appears because du = 3sin 3x dx, but you need sin 3x dx = (1/3)du, and the derivative of -cos 3x gives you +3sin 3x dx.

Watch out: Pay careful attention to signs when working with trigonometric functions - they can flip unexpectedly!

$--- OCR Start --- உ PAGE..........DATE... Calculus 2 - Integral Calculus - Anti -differentiation $\int f(x)dx = F(x) + C$ $\int$ integral si$

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Advanced Logarithmic Integration

Sometimes you'll encounter quotients involving more complex trigonometric expressions. For ∫(sec²4x dx)/(5tan 4x), factor out 1/5 first, then recognise that d/dx(tan 4x) = 4sec²4x.

Set u = tan 4x, so du = 4sec²4x dx, making sec²4x dx = (1/4)du. Your integral becomes (1/5) · (1/4)∫ $du/u$ = (1/20)ln|tan 4x| + C.

The absolute value in ln|u| is crucial because tangent can be negative, and you can't take the logarithm of negative numbers. This keeps your answer mathematically sound across all domains.

Final tip: Always include absolute value signs in logarithmic integration - it's a common exam mistake to forget them!

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Elisha

iOS user

Paul T

iOS user