S x²-x+1 x² + 2x + 2 2 -dx

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...
icon
Related questions
Question

Solve using the integration of rational function

08 - Rational Fractions.pdf
5
MADE4Learners
FRAMEWORK
Math138 - Engineering Calculus 2
1
How to resolve a proper rational fraction into its partial fractions.
If a linear factor ax + b occurs once as a factor of the
denominator, there corresponds to this factor a partial fraction
of the from
where A is a constant to be determined.
A
ax+b'
• If a linear factor ax + b occurs n times as a factor of the
denominator, there corresponds to this factor ʼn partial fractions
in the form
A₁
A₂
An
+
ax + b (ax + b)²
(ax + b)n
where A's are constants to be determined.
MADE4Learners
FRAMEWORK
Math138 - Engineering Calculus 2
2
● If an irreducible (not factorable) quadratic factor ax² + bx + c
occurs once as a factor in the denominator, there corresponds to
this factor one partial fraction in the form A(2ax+b)+B, where A
and B are constants to be determined and 2ax + b is the
derivative of ax² + bx + c.
ax²+bx+c ¹
● If an irreducible (not factorable) quadratic factor ax² + bx+c
occurs n times as a factor in the denominator, there corresponds
to this factor n partial fractions in the form
A₁ (2ax + b) + B₁,
ax² + bx + c
+
A₂(2ax + b) + B₂
(ax² + bx + c)²
+
An(2ax + b) + Bn
(ax²+bx+c)n
where A's and Bi's are constants to be determined and 2ax + b
is the derivative of ax² + bx + c.
3
Sac
Math138 - Engineering Calculus 2
3
How to solve the constants?
✪ By assigning a particular value to x so that some of the
Transcribed Image Text:08 - Rational Fractions.pdf 5 MADE4Learners FRAMEWORK Math138 - Engineering Calculus 2 1 How to resolve a proper rational fraction into its partial fractions. If a linear factor ax + b occurs once as a factor of the denominator, there corresponds to this factor a partial fraction of the from where A is a constant to be determined. A ax+b' • If a linear factor ax + b occurs n times as a factor of the denominator, there corresponds to this factor ʼn partial fractions in the form A₁ A₂ An + ax + b (ax + b)² (ax + b)n where A's are constants to be determined. MADE4Learners FRAMEWORK Math138 - Engineering Calculus 2 2 ● If an irreducible (not factorable) quadratic factor ax² + bx + c occurs once as a factor in the denominator, there corresponds to this factor one partial fraction in the form A(2ax+b)+B, where A and B are constants to be determined and 2ax + b is the derivative of ax² + bx + c. ax²+bx+c ¹ ● If an irreducible (not factorable) quadratic factor ax² + bx+c occurs n times as a factor in the denominator, there corresponds to this factor n partial fractions in the form A₁ (2ax + b) + B₁, ax² + bx + c + A₂(2ax + b) + B₂ (ax² + bx + c)² + An(2ax + b) + Bn (ax²+bx+c)n where A's and Bi's are constants to be determined and 2ax + b is the derivative of ax² + bx + c. 3 Sac Math138 - Engineering Calculus 2 3 How to solve the constants? ✪ By assigning a particular value to x so that some of the
S
x²-x+1
x² + 2x + 2
-dx
Transcribed Image Text:S x²-x+1 x² + 2x + 2 -dx
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question

Can you solve using the same method shown on the first picture please

2x-10
dx
= ( ( + = 1/5 ) dx
X-3
= 12/1dx - 12/²3 dx
= In
• 3 ³ /x + ₁] - √n / x - 3] + C ]
2x-10
(x+1)(x-3)
2x-10
(X+1)(x-3)
2x-10
Let x= -1
->
A
B
t
X + 1
X-3
A (x-3) + B(x+1)
(x+1) (x-3)
A (X-3) + B(x+1)
2(-1)-10 = A (-1-3) + B (-1+1)
-12-4 A
A = 3
H
Let x = 3
2 (3)-10= A (3-3) + B (3-1)
- 4 = 48
B = -.
Transcribed Image Text:2x-10 dx = ( ( + = 1/5 ) dx X-3 = 12/1dx - 12/²3 dx = In • 3 ³ /x + ₁] - √n / x - 3] + C ] 2x-10 (x+1)(x-3) 2x-10 (X+1)(x-3) 2x-10 Let x= -1 -> A B t X + 1 X-3 A (x-3) + B(x+1) (x+1) (x-3) A (X-3) + B(x+1) 2(-1)-10 = A (-1-3) + B (-1+1) -12-4 A A = 3 H Let x = 3 2 (3)-10= A (3-3) + B (3-1) - 4 = 48 B = -.
S
x²-x+1
x² + 2x + 2
-dx
Transcribed Image Text:S x²-x+1 x² + 2x + 2 -dx
Solution
Bartleby Expert
SEE SOLUTION
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning