26.5 Partial Fractions Problem Set

PARTIAL FRACTIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: Integrate the following function:

𝑓 𝑥 =

1 𝑥% − 4

(

./%

)

.0%

A. − ln ( B. −4ln (

./% .0%

)

(

./(

)

.0%

C. − ln (

)

)

(

D. − ln (𝑥) 2

SOLUTION 1: The TOPIC of PARTIAL FRACTIONS can be referenced under the topic of MATHEMATICS on page 28 and 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Integrating the function: 1 𝑑𝑥 𝑥% − 4 Made with

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The first step is to factor the denominator as much as possible and get the terms of the partial fraction decomposition using the table: 1 𝐴 𝐵 = + (𝑥 + 2)(𝑥 − 2) (𝑥 + 2) (𝑥 − 2) The next step is to add the right side back up by setting the terms together with a common denominator: 1 𝐴 𝑥 − 2 + 𝐵(𝑥 + 2) = (𝑥 + 2)(𝑥 − 2) (𝑥 + 2)(𝑥 − 2) Now, choose 𝐴 𝑎𝑛𝑑 𝐵 so that the numerators on each side are equal for every “𝑥". To do this, set the numerators equal: 1 = 𝐴 𝑥 − 2 + 𝐵(𝑥 + 2) Look for values that make one of the unknowns equal to zero and solve for the other. 𝑥 = −2: 1 = 𝐴 −2 − 2 + 𝐵 −2 + 2 ⇒ 𝐴 = − 𝑥 = 2: 1 = 𝐴 2 − 2 + 𝐵 2 + 2 ⇒ 𝐵 =

( )

( )

Plugging these values back in to the original decomposed function and carry out the integration. 1 1 4 + 4 )𝑑𝑥 ( 𝑥+2 𝑥−2 Made with −

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1 1 − 𝑙𝑛 𝑥 + 2 + ln 𝑥 − 2 + 𝐶 4 4 Recall that ln 𝑚 − ln 𝑛 = 𝑙𝑛

@ A

, so the solution is:

1 1 𝑥+2 𝑑𝑥 = − ln ( ) 𝑥% − 4 4 𝑥−2 𝟏

𝒙/𝟐

𝟒

𝒙0𝟐

Therefore, the correct answer choice is A. − 𝒍𝒏 (

PROBLEM 2: Evaluate the integral: 𝑥% + 𝑥 − 1 𝑑𝑥 𝑥(𝑥 % − 1) (

(.0()

%

(./()

B. ln 𝑥 + ln

(

(.0()

%

(./()

C. ln 𝑥 + 2ln

(.0()

A. ln 𝑥 + ln

(

.0(

%

./(

D. ln

(./()

+𝐶

+𝐶

+𝐶

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)

SOLUTION 2: The TOPIC of PARTIAL FRACTIONS can be referenced under the topic of MATHEMATICS on page 28 and 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The first step is to factor the denominator as much as possible and get the terms of the partial fraction decomposition using the table: 𝑥% + 𝑥 − 1 𝑥% + 𝑥 − 1 𝐴 𝐵 𝐶 = = + + 𝑥(𝑥 % − 1) 𝑥(𝑥 + 1)(𝑥 − 1) 𝑥 (𝑥 + 1) (𝑥 − 1) The next step is to add the right side back up by setting the terms together with a common denominator: 𝑥% + 𝑥 − 1 𝐴 𝑥 + 1 𝑥 − 1 + 𝐵𝑥 𝑥 − 1 + 𝐶𝑥(𝑥 + 1) = 𝑥(𝑥 + 1)(𝑥 − 1) 𝑥(𝑥 + 1)(𝑥 − 1)

Now, choose 𝐴, 𝐵, 𝑎𝑛𝑑 𝐶 so that the numerators on each side are equal for every “𝑥”. To do this, set the numerators equal: 𝑥 % + 𝑥 − 1 = 𝐴 𝑥 + 1 𝑥 − 1 + 𝐵𝑥 𝑥 − 1 + 𝐶𝑥(𝑥 + 1) Look for values that make one of the unknowns equal to zero and solve for the others. 𝑥 = 0: 0% + 0 − 1 = 𝐴 0 + 1)(0 − 1 + 𝐵 0)(0 − 1 + 𝐶(0)(0 + 1) ⇒ 𝐴 = 1 𝑥 = −1: −1% − 1 + 1 = 𝐴 −1 + 1 −1 − 1 + 𝐵 −1)(0 − 1 + 𝐶 −1 −1 + 1 ⇒ 𝐵 = −

( %

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𝑥 = 1: 1% + 1 − 1 = 𝐴 1 + 1 (1 − 1) + 𝐵 1)(1 − 1 + 𝐶(1)(1 + 1) ⇒ 𝐶 =

( %

Plugging these values back in to the original decomposed function and carry out the integration.

1 1 − 1 2 + 2 + 𝑑𝑥 𝑥 𝑥+1 𝑥−1

1 1 ln 𝑥 = − ln 𝑥 + 1 + ln 𝑥 − 1 + 𝐶 2 2 Recall that ln 𝑚 − ln 𝑛 = 𝑙𝑛

@ A

, so the solution is:

𝑥% + 𝑥 − 1 1 (𝑥 − 1) 𝑑𝑥 = ln 𝑥 + ln +𝐶 𝑥(𝑥 % − 1) 2 (𝑥 + 1) 𝟏

(𝒙0𝟏)

𝟐

(𝒙/𝟏)

Therefore, the correct answer choice is A. 𝒍𝒏 𝒙 + 𝒍𝒏

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+𝑪

PROBLEM 3: Evaluate the integral:

𝑥+7 𝑑𝑥 𝑥 % (𝑥 + 2)

2

A. − ln 𝑥 + 𝐶 ) 2

L

)

%.

2

L

)

%.

B. − ln 𝑥 − C. − ln 𝑥 − D. −

L %.

2

+ ln 𝑥 + 2 + 𝐶 )

2

+ ln 𝑥 + 2 + 𝐶 )

SOLUTION 3: The TOPIC of PARTIAL FRACTIONS can be referenced under the topic of MATHEMATICS on page 28 and 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The first step is to factor the denominator as much as possible, which in this case it already is, and get the terms of the partial fraction decomposition using the table.

𝑥+7 𝐴 𝐵 𝐶 = + + 𝑥 % (𝑥 + 2) 𝑥 𝑥 % (𝑥 + 2)

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The next step is to add the right side back up by setting the terms together with a common denominator:

𝑥+7 𝐴 𝑥 𝑥 + 2 + 𝐵 𝑥 + 2 + 𝐶 𝑥% = 𝑥% 𝑥 + 2 𝑥% 𝑥 + 2

Now, choose 𝐴, 𝐵, 𝑎𝑛𝑑 𝐶 so that the numerators on each side are equal for every “𝑥”. To do this, set the numerators equal: 𝑥 + 7 = 𝐴 𝑥 𝑥 + 2 + 𝐵 𝑥 + 2 + 𝐶(𝑥 % ) Look for values that make one of the unknowns equal to zero and solve for the others. 𝑥 = 0: 0 + 7 = 𝐴(0)(0 + 2) + 𝐵 0 + 2 + 𝐶(0% ) ⇒ 𝐵 =

L %

𝑥 = −1: − 1 + 7 = 𝐴 −1)(−1 + 2 + 𝐵 −1 + 2 + 𝐶 −1% ⇒ 𝐶 = 𝑥 = 1: 1 + 7 = 𝐴 1 (1 + 2) + 𝐵 1 + 2 + 𝐶(−1% ) ⇒ 𝐴 =

2 )

0( )

Plugging these values back in to the original decomposed function and carry out the integration.

5 7 5 4 − 4 + 2% − 𝑑𝑥 𝑥 𝑥 (𝑥 + 2) 5 7 5 − ln 𝑥 − + ln 𝑥 + 2 + 𝐶 4 2𝑥 4 Made with

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So the solution is:

𝑥+7 5 7 5 𝑑𝑥 = − ln 𝑥 − + ln 𝑥 + 2 + 𝐶 𝑥 % (𝑥 + 2) 4 2𝑥 4

Therefore, the correct answer choice is 𝟓

𝟕

𝟒

𝟐𝒙

C. − 𝒍𝒏 𝒙 −

𝟓

+ 𝒍𝒏 𝒙 + 𝟐 + 𝑪 𝟒

PROBLEM 4: Evaluate the integral: 4 𝑑𝑥 𝑥 % + 5𝑥 − 14 )

)

P

P

A. 𝑙𝑛 𝑥 − 2 − 𝑙𝑛 𝑥 + 7 + 𝐶 )

B. 𝑙𝑛 𝑥 − 2 + 7 P

)

)

P

P

C. 𝑙𝑛 𝑥 − 2 + 𝑙𝑛 𝑥 + 7 + 𝐶 )

D. 𝑙𝑛 𝑥 + 22 P

SOLUTION 4: The TOPIC of PARTIAL FRACTIONS can be referenced under the topic of MATHEMATICS on page 28 and 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

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To get the problem started off we need the form of the partial fraction decomposition of the integrand. However, in order to get this we’ll need to factor the denominator. 4 𝑑𝑥 = 𝑥 % + 5𝑥 − 14

4 𝑑𝑥 (𝑥 + 7)(𝑥 − 2)

The form of the partial fraction decomposition for the integrand is then, 4 𝐴 𝐵 = + 𝑥 + 7 (𝑥 − 2) (𝑥 + 7) (𝑥 − 2)

Setting the numerators equal gives, 4=𝐴 𝑥−2 +𝐵 𝑥+7 Look for values that make one of the unknowns equal to zero and solve for the others.

𝑥 = 2: 4 = 𝐴(2 − 2) + 𝐵 2 + 7 ⇒ 𝐵 =

4 9

𝑥 = −7: 4 = 𝐴 −7 − 2 + 𝐵 −7 + 7 ⇒ 𝐴 = −

4 9

The partial fraction form of the integrand is then, 4 4 − 4 9 + 9 = 𝑥 + 7 (𝑥 − 2) (𝑥 + 7) (𝑥 − 2)

We can now evalaute the integral. Made with

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4 𝑑𝑥 = (𝑥 + 7)(𝑥 − 2)

4 9 𝑑𝑥 + (𝑥 + 7) −

4 4 4 9 𝑑𝑥 = 𝑙𝑛 𝑥 − 2 − 𝑙𝑛 𝑥 + 7 + 𝐶 (𝑥 − 2) 9 9

Therefore, the correct answer choice is 𝟒

𝟒

𝟗

𝟗

A. 𝒍𝒏 𝒙 − 𝟐 − 𝒍𝒏 𝒙 + 𝟕 + 𝑪 PROBLEM 5: Evaluate the integral: 3𝑥 + 11 𝑑𝑥 𝑥% − 𝑥 − 6 A. 𝑙𝑛 𝑥 − 3 − 𝑙𝑛 𝑥 + 2 B. 𝑙𝑛 𝑥 − 3 − 2𝑙𝑛 𝑥 + 2 + 𝐶 C. 𝑙𝑛 𝑥 + 1 D. 4𝑙𝑛 𝑥 − 3 − 𝑙𝑛 𝑥 + 2 + 𝐶

SOLUTION 5: The TOPIC of PARTIAL FRACTIONS can be referenced under the topic of MATHEMATICS on page 28 and 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives, 3𝑥 + 11 𝑑𝑥 = 𝑥% − 𝑥 − 6

3𝑥 + 11 𝐴 𝐵 𝑑𝑥 = + (𝑥 − 3)(𝑥 + 2) (𝑥 − 3) (𝑥 + 2)

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The next step is to actually add the right side back up. 3𝑥 + 11 𝐴(𝑥 + 2) 𝐵(𝑥 − 3) = + 𝑥 − 3 (𝑥 + 2) (𝑥 − 3) (𝑥 + 2)

Setting the numerators equal gives, 3𝑥 + 11 = 𝐴 𝑥 + 2 + 𝐵 𝑥 − 3 Look for values that make one of the unknowns equal to zero and solve for the others. 𝑥 = −2: 5 = 𝐴 −2 + 2 + 𝐵 −2 − 3 ⇒ 𝐵 = −1 𝑥 = 3: 20 = 𝐴 3 + 2 + 𝐵 3 − 3 ⇒ 𝐴 = 4

The partial fraction form of the integrand is then, 3𝑥 + 11 4 1 = − 𝑥 − 3 (𝑥 + 2) (𝑥 − 3) (𝑥 + 2)

We can now do the integral. 3𝑥 + 11 𝑑𝑥 = (𝑥 − 3)(𝑥 + 2)

4 𝑑𝑥 − (𝑥 − 3)

1 𝑑𝑥 = 4𝑙𝑛 𝑥 − 3 − 𝑙𝑛 𝑥 + 2 + 𝐶 (𝑥 + 2)

Therefore, the correct answer choice is D. 𝟒𝒍𝒏 𝒙 − 𝟑 − 𝒍𝒏 𝒙 + 𝟐 + 𝑪

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