The integral dæ appears on our list of basic antiderivatives. What is the result? arctan(x) + C In]1+ a*| + C_ Question 2 The integrand in the problem dx can be re-written as a product, like this: 1+x² Sz- () da . 1+z 1 To apply integration by parts, we can let u = x, and dv = 1+z² Then du = 1dx , and v = arctan x . What are the next steps for trying the integration by parts technique here? uv = fv du *· arctan r – J læ dx = ¤ • arctan x –x² + C uv – fv du = ¤· arctan ¤ – f arctan x dx Stop here and try a different approach, because we don't have a good way to antidifferentiate arctan: uv – fv du = ¤•arctan ¤ – 1+z² = x • arctan x – arctanx +C
The integral dæ appears on our list of basic antiderivatives. What is the result? arctan(x) + C In]1+ a*| + C_ Question 2 The integrand in the problem dx can be re-written as a product, like this: 1+x² Sz- () da . 1+z 1 To apply integration by parts, we can let u = x, and dv = 1+z² Then du = 1dx , and v = arctan x . What are the next steps for trying the integration by parts technique here? uv = fv du *· arctan r – J læ dx = ¤ • arctan x –x² + C uv – fv du = ¤· arctan ¤ – f arctan x dx Stop here and try a different approach, because we don't have a good way to antidifferentiate arctan: uv – fv du = ¤•arctan ¤ – 1+z² = x • arctan x – arctanx +C
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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Transcribed Image Text:The integral J
dx
1+z²
appears on our list of basic antiderivatives.
What is the result?
arctan(x) + C
+C
In|1 + æ*| + C
Question 2
The integrand in the problem f ,dx can be re-written as a product, like this:
1+z?
Jn. (규E) de.
1+x²
To apply integration by parts, we can let u = x, and dv =
1+z²
Then du = 1dx , and v = arctan x .
What are the next steps for trying the integration by parts technique here?
Sv du
= x • arctan æ – S 1æ dx
Uv =
* · arctan a – x² + C
uv – fv du
= x · arctan ¤ – farctan æ dæ
Stop here and try a different approach, because we don't have a good way to antidifferentiate arctan x .
uv – Sv du
= ¤· arctan ¤ – S dx
1+z²
= x • arctan x – arctan x + C
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