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#### The Fundamental Theorem of Calculus

Three Different Quantities
The Whole as Sum of Partial Changes
The Indefinite Integral as Antiderivative
The FTC and the Chain Rule

#### The Indefinite Integral and the Net Change

Indefinite Integrals and Anti-derivatives
A Table of Common Anti-derivatives
The Net Change Theorem
The NCT and Public Policy

#### Substitution

Substitution for Indefinite Integrals
Revised Table of Integrals
Substitution for Definite Integrals

#### Area Between Curves

The Slice and Dice Principle
To Compute a Bulk Quantity
The Area Between Two Curves
Horizontal Slicing
Summary

#### Volumes

Slicing and Dicing Solids
Solids of Revolution 1: Disks
Solids of Revolution 2: Washers
Volumes Rotating About the $y$-axis

Behind IBP
Examples
Going in Circles

#### Integrals of Trig Functions

Basic Trig Functions
Product of Sines and Cosines (1)
Product of Sines and Cosines (2)
Product of Secants and Tangents
Other Cases

#### Trig Substitutions

How it works
Examples
Completing the Square

#### Partial Fractions

Introduction
Linear Factors
Improper Rational Functions and Long Division
Summary

#### Strategies of Integration

Substitution
Integration by Parts
Trig Integrals
Trig Substitutions
Partial Fractions

#### Improper Integrals

Type I Integrals
Type II Integrals
Comparison Tests for Convergence

#### Differential Equations

Introduction
Separable Equations
Mixing and Dilution

#### Models of Growth

Exponential Growth and Decay
Logistic Growth

#### Infinite Sequences

Close is Good Enough (revisited)
Examples
Limit Laws for Sequences
Monotonic Convergence

#### Infinite Series

Introduction
Geometric Series
Limit Laws for Series
Telescoping Sums and the FTC

#### Integral Test

The Integral Test
When the Integral Diverges
When the Integral Converges

#### Comparison Tests

The Basic Comparison Test
The Limit Comparison Test

#### Convergence of Series with Negative Terms

Introduction
Alternating Series and the AS Test
Absolute Convergence
Rearrangements

The Ratio Test
The Root Test
Examples

#### Strategies for testing Series

List of Major Convergence Tests
Examples

#### Power Series

Finding the Interval of Convergence
Other Power Series

#### Representing Functions as Power Series

Functions as Power Series
Derivatives and Integrals of Power Series
Applications and Examples

#### Taylor and Maclaurin Series

The Formula for Taylor Series
Taylor Series for Common Functions
Adding, Multiplying, and Dividing Power Series
Miscellaneous Useful Facts

#### Applications of Taylor Polynomials

What are Taylor Polynomials?
How Accurate are Taylor Polynomials?
What can go Wrong?
Other Uses of Taylor Polynomials

#### Partial Derivatives

Definitions and Rules
The Geometry of Partial Derivatives
Higher Order Derivatives
Differentials and Taylor Expansions

#### Multiple Integrals

Background
What is a Double Integral?
Volumes as Double Integrals

#### Iterated Integrals over Rectangles

One Variable at the Time
Fubini's Theorem
Notation and Order

#### Double Integrals over General Regions

Type I and Type II regions
Examples
Order of Integration
Area and Volume Revisited

### Linear Factors

 For every factor of $(x-a)$ in $Q(x)$, we have a term $\displaystyle\frac{A}{x-a}$. For every repeated linear factor $(x-a)^n$, we have $$\displaystyle\frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2} + \ldots + \frac{A_n}{(x-a)^n}.$$

The video below goes over some examples.

#### More Examples

 Example 1: Find $\displaystyle\int\frac6{x^2-1}\,dx$. Solution: Since $x^2-1= (x-1)(x+1)$, we look for a decomposition of the form $$\displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ { A \over x-1 } + { B \over x+1 } }$$ for some $A,\,B$. Taking the common denominator, we have $$\displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ { A(x+1) +B(x-1)\over (x-1)(x+1) }}.$$ This is true if the numerators are equal, i.e., $$6=A(x+1)+B(x-1).$$ To find out $A$, substitute $x=1$, then $6=A(2)+B(0)$, so $A=3$. Similarly, substituting $x=-1$ lets us find that $B=-3$. Finally, we have $$\int \displaystyle{ 6 \over (x-1)(x+1) }\, dx= \int \displaystyle{ {3 \over x-1}\,dx +\int {-3 \over x+1}\,dx }=3\log{(x-1)}-3\log{(x+1)}+ C.$$

 Example 2: Find the partial fraction decomposition of $$\frac{1}{x^4-3x^3+3x^2-x}.$$ Solution: Since $Q(x)=x\,\left(x-1\right)^3$, we have both a non-repeated factor, $x$, and a repeated factor, $(x-1)^3$, so we're looking for a decomposition of the form $$\frac{1}{x\,\left(x-1\right)^3}=\frac{A}{x}+\frac{B_1}{x-1}+\frac{B_2}{\left(x-1\right)^2}+\frac{B_3}{\left(x-1\right)^3}.$$ Taking the common denominator, we have $$\frac{1}{x\,\left(x-1\right)^3}=\frac{A\left(x-1\right)^3+B_1x\,(x-1)^2+B_2x\,\left(x-1\right)+B_3x} {x\left(x-1\right)^3}.$$ This is true if the numerators are equal, i.e., $$1=A\left(x-1\right)^3+B_1x\,(x-1)^2+B_2x\,\left(x-1\right)+B_3x.$$ To find out $B_3$, substitute $x=1$, then $1=0+B_3$, so $B_3=1$. Similarly, substituting $x=0$ let us find that $A=-1$. To find the other coefficients, we need to choose other values for $x$. For example, taking $x=2$ yields $$1=A+2B_1+2B_2+2B_3=-1+2B_1+2B_2+2,$$ i.e., $$B_1+B_2=0.$$ Taking $x=-1$ yields $$1=-8A-4B_1+2B_2-B_3=8-4B_1+2B_2-1,$$ i.e., $$-2B_1+B_2=-6.$$ Solving the system $$\begin{cases} B_1+B_2=0\\ -2B_1+B_2=-6 \end{cases}$$ yields $B_1=2$ and $B_2=-2$, so that our decomposition becomes $$\frac{1}{x\,\left(x-1\right)^3}=-\frac{1}{x}+\frac{2}{x-1}-\frac{2}{\left(x-1\right)^2}+\frac{1}{\left(x-1\right)^3}.$$