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Integrate 1/x2+a2.dx=1

  • Thread starter king imran
  • Start date
integratio 1/x2+a2.dx=1
intervel0 to +infinity
find value of 'a'
 

Answers and Replies

Help Me Inthis Question
 
Hootenanny
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Hi there imran,

I'm guessing that your integrand, is something along the lines of;

[tex]\int^{\infty}_{0} \frac{1}{x^2} + a^2 \;dx = 1[/tex]

Please note, that the rules of this forum prohibit us to provide help to someone who has not attempted there homework question before hand (assuming of course this is homework). Perhaps, if you posted your attempt, we could help you further...
 
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The a^2 factor must be on the denominator, otherwise the integral is divergent both horizontally at infinity and vertically at zero.

Just clearing up the interpretation. Since the original poster has access to the original problem anyway, this doesn't constitute helping him. ;)

--Stuart Anderson
 
Gib Z
Homework Helper
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The Integral [tex]\int \frac{1}{x^2 + a^2} dx[/tex] looks very familiar to me :)

Maybe some trig will help you out?
 
Hootenanny
Staff Emeritus
Science Advisor
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9,598
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The a^2 factor must be on the denominator, otherwise the integral is divergent both horizontally at infinity and vertically at zero.

Just clearing up the interpretation. Since the original poster has access to the original problem anyway, this doesn't constitute helping him. ;)

--Stuart Anderson
You are quite correct of course!
The Integral [tex]\int \frac{1}{x^2 + a^2} dx[/tex] looks very familiar to me :)
And so it should :wink:
Maybe some trig will help you out?
Good suggestion :approve:
 
find volume of solid y=x2-1 ,x=2 ,y=0 is relvived about y-asis
 
HallsofIvy
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If this is a new problem, start a new thread.

Also, since it is obviously homework, post it in the homework thread and show what you have tried!
 
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plz give me good sites to refer for integration,its applications
 
3
0
[tex]\[
\int {\frac{1}
{{x^2 + a^2 }}dx} = \frac{1}
{a}\tan ^{ - 1} \frac{x}
{a} + c
\][/tex]

Just in case.
 
Hootenanny said:
[tex]\int^{\infty}_{0} \frac{1}{x^2} + a^2 \;dx = 1[/tex]
does the following integral even exist?
[tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]
 
cristo
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does the following integral even exist?
[tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]
Look at the above post:

The a^2 factor must be on the denominator, otherwise the integral is divergent both horizontally at infinity and vertically at zero.

Just clearing up the interpretation. Since the original poster has access to the original problem anyway, this doesn't constitute helping him. ;)

--Stuart Anderson
 
Gib Z
Homework Helper
3,344
4
does the following integral even exist?
[tex]\int^{\infty}_{0}\left(\frac{1}{x^2} + a^2\right) dx[/tex]
The question was actually with the a^2 also in the denominator, look at above posts. And that integral you stated only exists if a=0.
 

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