Partial Fractions

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Substitution for Indefinite Integrals
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To Compute a Bulk Quantity
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$y$ -axis

Integration by Parts

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Partial Fractions

Introduction
Linear Factors
Quadratic 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
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Close is Good Enough (revisited)
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Convergence of Series with Negative Terms

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The Ratio and Root Tests

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Strategies for testing Series

List of Major Convergence Tests
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Power Series

Radius and Interval of Convergence
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Representing Functions as Power Series

Functions as Power Series
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Partial Derivatives

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Higher Order Derivatives
Differentials and Taylor Expansions

Multiple Integrals

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Volumes as Double Integrals

Iterated Integrals over Rectangles

One Variable at the Time
Fubini's Theorem
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Double Integrals over General Regions

Type I and Type II regions
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Area and Volume Revisited

A summary

Here are the steps for computing $\displaystyle\int \frac{P(x)}{Q(x)}\,dx$ , where $P(x)$ and $Q(x)$ are polynomials.

Check that we have a proper fraction. If not, do long division.

Factor $Q(x)$ into a product of linear and irreducible quadratic terms.

Write P(x)Q(x) as a sum of terms with unknown coefficients:
- 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 $\displaystyle\frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2} + \ldots + \frac{A_n}{(x-a)^n}.$
- For every quadratic factor $x^2 + bx + c$ , we have $\displaystyle\frac{Ax+B}{x^2+bx+c}.$
- For every repeated quadratic factor $(x^2+bx+c)^n$ , we have $\displaystyle\frac{A_1 x + B_1}{x^2+bx+c} + \frac{A_2x+B_2}{(x^2+bx+c)^2}+\ldots + \frac{A_nx+B_n}{(x^2+bx+c)^n}.$

Multiply both sides of the equation by $Q(x)$ . Then plug in different values of $x$ to get equations that determine the coefficients. For distinct linear factors, we can always get the coefficient of $\displaystyle\frac{1}{x-a}$ by plugging in $x=a$ . For repeated linear factors or for quadratic factors, it can be a little trickier.

If all else fails, compare the coefficients of 1, $x$ , $x^2$ , etc. on both sides to get equations involving $A$ , $B$ , etc.

Integrate. Recall that
- $\int \frac{1}{x-a}\,dx =\ln\lvert x-a\rvert +C$
- $\int\frac{1}{(x-a)^n}\,dx =\frac{1}{1-n}(x-a)^{1-n}+C$
- $\int\frac{1}{x^2+a^2}\,dx=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) +C$
- $\int\frac{x }{x^2+a^2}\,dx =\frac{1}{2}\ln\left(x^2+a^2\right)+C$
- Integrating other quadratic expressions is done with either a $u$ -substitution or a trig substitution $x=a\tan(\theta)$ , after possibly completing the square.