<< Prev Next >>


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

Area Between Curves

Computation Using Integration
To Compute a Bulk Quantity
The Area Between Two Curves
Horizontal Slicing


Slicing and Dicing Solids
Solids of Revolution 1: Disks
Solids of Revolution 2: Washers
More Practice

Integration by Parts

Integration by Parts
Integration by Parts with a definite integral
Going in Circles
Tricks of the Trade

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
Completing the Square

Partial Fractions

Linear Factors
Irreducible Quadratic Factors
Improper Rational Functions and Long Division

Strategies of Integration

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

Separable Equations
Mixing and Dilution

Models of Growth

Exponential Growth and Decay
Logistic Growth

Infinite Sequences

Approximate Versus Exact Answers
Examples of Infinite Sequences
Limit Laws for Sequences
Theorems for and Examples of Computing Limits of Sequences
Monotonic Covergence

Infinite Series

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

Integral Test

Road Map
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

The Ratio and Root Tests

The Ratio Test
The Root Test

Strategies for testing Series

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

Power Series

Radius and Interval of Convergence
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

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


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 $$ \frac{1}{x\,\left(x-1\right)^3}=-\frac{1}{x}+\frac{2}{x-1}-\frac{2}{\left(x-1\right)^2}+\frac{1}{\left(x-1\right)^3}. $$

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. 

  Differentiate to see if this answer is right.