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
Your answer is wrong plz give me this answer i attached
With steps
Transcribed Image Text:5:18
//V2
Solution
bovz.
X
1 + 4x
X -dx
1+1x¹
X 2-4arctan (√2+1) + 4arctan (√2-1)
√√2
16
1+ 4x
(Decimal: 0.19634...)
Steps
dx
1(2x²-
We
Take the partial fraction of
1
- 2x + 1
4(2x².
dx
4
1
(2x²+2x+1
4(2x²
1/2x²+2x+1)+1/2²2²-²23-1²x
Solution
1
=
X
+2x+1)
1
1+ 4x
Apply the Sum Rule: ff(x) + g(x)dx=ff(x) dx = fg(x) dx
-1/²
67.
-dx + Jov
Show Steps ✓
4G, 53%
1
4(2x²+2x+1)
-dx = =|arctan(v2 + 1) -
1
4(2x²–2x+1)
Show Steps
-dx
dx = (arctan (√2-1) + )
4(2x² 2x+1)
+
Show Steps
Show Steps
![5:18
S
1+ 4x
00:20
also,
putting
Then we get,
1152
_|||
1+ 4x
=
2x² = t
=) 2 (2x) dx = dt
s dx
So
when x=0, + = 0
2
when x= 1/², += x (x₁) =
(15₂)
√2
dx =
therefore,
152
% 1+ (22²)
16
O
= + (1-0)
1+ 4x
dt
**
= = (tan²¹ (+))"
= = [tan^¹ (₁) - tan²¹ (0)]
4
dx =
dt
4
?
(dd/4)
O
1+ t
+ +1
π
16
4G 53%
Final Answer
√x
8
Do](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd16325ae-6837-4cb1-ac21-99766adab78f%2Ff5dda7a3-0035-4271-a945-da52ea1810a5%2Fb5epwui_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5:18
S
1+ 4x
00:20
also,
putting
Then we get,
1152
_|||
1+ 4x
=
2x² = t
=) 2 (2x) dx = dt
s dx
So
when x=0, + = 0
2
when x= 1/², += x (x₁) =
(15₂)
√2
dx =
therefore,
152
% 1+ (22²)
16
O
= + (1-0)
1+ 4x
dt
**
= = (tan²¹ (+))"
= = [tan^¹ (₁) - tan²¹ (0)]
4
dx =
dt
4
?
(dd/4)
O
1+ t
+ +1
π
16
4G 53%
Final Answer
√x
8
Do
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