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#### The Fundamental Theorem of Calculus

Three Different Concepts
The Fundamental Theorem of Calculus (Part 2)
The Fundamental Theorem of Calculus (Part 1)
More FTC 1

#### 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
Examples to Try
Revised Table of Integrals
Substitution for Definite Integrals
Examples

#### Area Between Curves

Computation Using Integration
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
More Practice

#### Integration by Parts

Integration by Parts
Examples
Integration by Parts with a definite integral
Going in Circles

#### Integrals of Trig Functions

Antiderivatives of Basic Trigonometric Functions
Product of Sines and Cosines (mixed even and odd powers or only odd powers)
Product of Sines and Cosines (only even powers)
Product of Secants and Tangents
Other Cases

#### Trig Substitutions

How Trig Substitution Works
Summary of trig substitution options
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 1 - Improper Integrals with Infinite Intervals of Integration
Type 2 - Improper Integrals with Discontinuous Integrands
Comparison Tests for Convergence

#### Differential Equations

Introduction
Separable Equations
Mixing and Dilution

#### Models of Growth

Exponential Growth and Decay
Logistic Growth

#### Infinite Sequences

Examples of Infinite Sequences
Limit Laws for Sequences
Theorems for and Examples of Computing Limits of Sequences
Monotonic Covergence

#### Infinite Series

Introduction
Geometric Series
Limit Laws for Series
Test for Divergence and Other Theorems
Telescoping Sums and the FTC

#### Integral Test

The Integral Test
Estimates of Value of the Series

#### 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

Strategy to Test Series and a Review of Tests
Examples, Part 1
Examples, Part 2

#### Power Series

Finding the Interval of Convergence
Power Series Centered at $x=a$

#### 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

Taylor Polynomials
When Functions Are Equal to Their Taylor Series
When a Function Does Not Equal Its Taylor Series
Other Uses of Taylor Polynomials

#### Partial Derivatives

Visualizing Functions in 3 Dimensions
Definitions and Examples
An Example from DNA
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

How To Compute Iterated Integrals
Examples of Iterated Integrals
Cavalieri's Principle
Fubini's Theorem
Summary and an Important Example

#### Double Integrals over General Regions

Type I and Type II regions
Examples 1-4
Examples 5-7
Order of Integration

### Linear Factors

Let $f(x)$ be the proper rational function $f(x)=\frac{P(x)}{Q(x)}$.  To use the method of partial fractions to integrate $f(x)$, we first factor $Q(x)$.  In the case when the factorization of $Q(x)$ contains linear factors, the following summary will help.

#### Linear factors of $Q(x)$

• For every factor of $(x-a)$ in $Q(x)$, we have a term $\displaystyle\frac{A}{x-a}$ in the decomposition.

• For every repeated linear factor $(x-a)^n$, we have the $n$ terms $$\displaystyle\frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2} + \ldots + \frac{A_n}{(x-a)^n}$$ in the decomposition.

The video below goes over some examples.

#### More Examples

Example 1: Find $\displaystyle\int\frac6{x^2-1}\,dx$.
DO:  Notice that substitution doesn't work here, but the trig substitution using secant will, and partial fractions will.  Try the problem both ways to see which you prefer.  The solution below uses partial fractions.  Don't read forward until you have tried the problem.

Solution 1: 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 $A$, 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 $$\int \displaystyle{ 6 \over (x-1)(x+1) }\, dx= \int \displaystyle{ {3 \over x-1}\,dx +\int {-3 \over x+1}\,dx }=3\log{\lvert x-1\rvert}-3\log{\lvert x+1\rvert}+ C.$$

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Example 2: Find the partial fraction decomposition of $$\frac{1}{x^4-3x^3+3x^2-x}.$$
DO:  The issue here is factoring a quartic, which can be quite difficult.  Hint:  $Q(x)=x\,\left(x-1\right)^3$.  Work on this before proceeding.

Solution 2: 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 $x=0$ we 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 $$\frac{1}{x\,\left(x-1\right)^3}=-\frac{1}{x}+\frac{2}{x-1}-\frac{2}{\left(x-1\right)^2}+\frac{1}{\left(x-1\right)^3}.$$
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Example 3:  DO:  Use your answer above to find $\displaystyle\int\frac{1}{x(x-1)^3}\,dx$.

Solution 3:  $\displaystyle\int\frac{1}{x\,\left(x-1\right)^3}\,dx=-\int\frac{1}{x}\,dx+\int\frac{2}{x-1}\,dx-\int\frac{2}{\left(x-1\right)^2}\,dx+\int\frac{1}{\left(x-1\right)^3}\,dx$

$\displaystyle=-\ln|x|+2\ln|x-1|+\frac{2}{(x-1)}-\frac{1}{2(x-1)^2}+C$,     using $u$-substitution on the last two terms.

DO:
Differentiate to see if this answer is right.