Use partial fractions to find the integral. S 4-x 2x²+x-1 Step 1 Rewrite the polynomial as 2x² + x1 = 2x² - x + 2x - 1. Collect the terms according to the power of x on the right side and factorize. 2x² + 2x -x-1= x(2x - 1) + (2x - 1) = (2x - 1)(x + 1) Because 2x² + x - 1 = (2x − 1)(x + 1), you should include one partial fraction for each factor and write B Step 2 4-x 2x²+x-1 dx where A and B are to be determined. 2x To solve for A, let x = 1/2✔ 4 2 Multiplying this equation by the least common denominator (2x - 1)(x + 1) yields the basic equation 4-x= A(x+1 + B(2x - 1). x+1 Because this equation is to be true for all x, substitute any convenient values for x to obtain equations in A and B. Step 3 The most convenient values are the ones that make particular factors equal to 0. Step 4 To solve for B, let x = 4 1/2 and obtain 4 4 − ½⁄2 = A( 1⁄2 + ¹) + B( 2 × ½1⁄2-1) 1x A = 7/3 ✔ 4 + 1 = A(-1 X =4 2x B = -5/3 2 1 7/3 X X X X to obtain the following. + 1) + B(2(-1 0 + B(4 1 x ) ✓ ) - 1)
Use partial fractions to find the integral. S 4-x 2x²+x-1 Step 1 Rewrite the polynomial as 2x² + x1 = 2x² - x + 2x - 1. Collect the terms according to the power of x on the right side and factorize. 2x² + 2x -x-1= x(2x - 1) + (2x - 1) = (2x - 1)(x + 1) Because 2x² + x - 1 = (2x − 1)(x + 1), you should include one partial fraction for each factor and write B Step 2 4-x 2x²+x-1 dx where A and B are to be determined. 2x To solve for A, let x = 1/2✔ 4 2 Multiplying this equation by the least common denominator (2x - 1)(x + 1) yields the basic equation 4-x= A(x+1 + B(2x - 1). x+1 Because this equation is to be true for all x, substitute any convenient values for x to obtain equations in A and B. Step 3 The most convenient values are the ones that make particular factors equal to 0. Step 4 To solve for B, let x = 4 1/2 and obtain 4 4 − ½⁄2 = A( 1⁄2 + ¹) + B( 2 × ½1⁄2-1) 1x A = 7/3 ✔ 4 + 1 = A(-1 X =4 2x B = -5/3 2 1 7/3 X X X X to obtain the following. + 1) + B(2(-1 0 + B(4 1 x ) ✓ ) - 1)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![Use partial fractions to find the integral.
4- X
| 2x²
2x² + x1
Step 1
Rewrite the polynomial as
2x² + x1 = 2x² - x + 2x - 1.
Collect the terms according to the power of x on the right side and factorize.
2x² + 2x -x-1 = x(2x − 1) + (2x - 1)
= (2x - 1)(x + 1)
Because 2x² + x − 1 = (2x − 1)(x + 1), you should include one partial fraction for each factor and write
A
4-X
2x²+x-1
where A and B are to be determined.
dx
To solve for A, let x = 1/2✔
1 x
2
4
2x
Step 2
Multiplying this equation by the least common denominator (2x - 1)(x + 1) yields the basic equation
4x = A(x+1
+ B(2x - 1).
Step 3
The most convenient values are the ones that make particular factors equal to 0.
7
x+1
Because this equation is to be true for all x, substitute any convenient values for x to obtain equations in A and B.
Step 4
To solve for B, let x = 4
1/2 and obtain
4
− —⁄2 = A ( ²¹1 + 1) + B( ² × 1⁄2 − 1)
2 -
2
A = 7/3
X
= AX
4 + 1 = A(-1
2x
= 4
B = -5/3
1
2
+ 5 x
7/3
x
X to obtain the following.
+ 1) + B(2(-1
B
X + B(4
X
× )
+ 1
✔)-1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0fdef185-de8b-42f5-933e-8018dcf36541%2F37d634e7-7204-4262-8b63-fa01f66dd5cc%2F1x2mtuuj_processed.png&w=3840&q=75)
Transcribed Image Text:Use partial fractions to find the integral.
4- X
| 2x²
2x² + x1
Step 1
Rewrite the polynomial as
2x² + x1 = 2x² - x + 2x - 1.
Collect the terms according to the power of x on the right side and factorize.
2x² + 2x -x-1 = x(2x − 1) + (2x - 1)
= (2x - 1)(x + 1)
Because 2x² + x − 1 = (2x − 1)(x + 1), you should include one partial fraction for each factor and write
A
4-X
2x²+x-1
where A and B are to be determined.
dx
To solve for A, let x = 1/2✔
1 x
2
4
2x
Step 2
Multiplying this equation by the least common denominator (2x - 1)(x + 1) yields the basic equation
4x = A(x+1
+ B(2x - 1).
Step 3
The most convenient values are the ones that make particular factors equal to 0.
7
x+1
Because this equation is to be true for all x, substitute any convenient values for x to obtain equations in A and B.
Step 4
To solve for B, let x = 4
1/2 and obtain
4
− —⁄2 = A ( ²¹1 + 1) + B( ² × 1⁄2 − 1)
2 -
2
A = 7/3
X
= AX
4 + 1 = A(-1
2x
= 4
B = -5/3
1
2
+ 5 x
7/3
x
X to obtain the following.
+ 1) + B(2(-1
B
X + B(4
X
× )
+ 1
✔)-1)
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