Partial Fraction Decomposition in Integration

This video tutorial introduces partial fraction decomposition for integration, explaining its advantages over trigonometric substitution. It provides two examples of integrating rational functions using partial fraction decomposition with linear factors. The video concludes with a preview of the next tutorial, which will cover quadratic factors.

What is the main advantage of using partial fraction decomposition in integration?

• It simplifies the integrand to avoid complex methods.

• It always results in a polynomial.

• It eliminates the need for any substitution.

• It is only applicable to trigonometric functions.

In the first example, what are the linear factors of the denominator?

• x - 3 and x + 2

• x + 3 and x - 2

• x - 4 and x + 1

• x + 4 and x - 1

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CCSS.HSA.APR.D.6

How do you determine the constants A and B in partial fraction decomposition?

• By solving a system of equations derived from the original equation.

• By guessing and checking.

• By integrating directly.

• By using the quadratic formula.

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CCSS.HSF-IF.C.7C

What is the purpose of letting x equal the zeros of the binomial factors?

• To eliminate the need for integration.

• To find the roots of the polynomial.

• To simplify the equation and solve for constants.

• To check the validity of the factors.

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CCSS.HSF-IF.C.7C

In the second example, what are the linear factors of the denominator?

• x - 4 and x + 2

• x + 3 and x - 1

• x + 4 and x - 2

• x - 3 and x + 1

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CCSS.HSF.IF.A.2

What is the value of B when x = -2 in the second example?

• 1

• -2

• -1

• 2

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CCSS.HSF.IF.A.2

What is the value of A when x = 4 in the second example?

• 2

• 1

• 0

• 4

What is the integral of 1 over U in terms of U?

• Natural log of the absolute value of U

• U squared

• 1 over U

• Exponential of U

Why is partial fraction decomposition useful in integration?

• It breaks down complex fractions into simpler ones.

• It is only applicable to trigonometric functions.

• It always results in a polynomial.

• It eliminates the need for any substitution.

What is the final form of the integral after applying partial fraction decomposition in the second example?

• ln|x - 4| + 4 * ln|x + 2| + C

• 4 * ln|x - 4| + ln|x + 2| + C

• 4 * ln|x + 4| + ln|x - 2| + C

• ln|x + 4| + 4 * ln|x - 2| + C