How to Integrate [1/(x^2 + 1)] dx?


by optics.tech
Tags: integration
optics.tech
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#1
Aug6-09, 02:45 PM
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Hi everyone,

Can you tell me how to integrate the following equation?

[tex]\int\frac{1}{x^2 + 1} \ dx[/tex]

I've tried the substitution method, u = x^2 + 1, du/dx = 2x. But the x variable is still exist.

Also, the trigonometry substitution method, but the denominator is not in [tex]\sqrt{x^2 + 1}[/tex] form.

Thanks in advance

Huygen
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tiny-tim
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#2
Aug6-09, 04:44 PM
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Hi Huygen!
Try x = tanu.
Hurkyl
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#3
Aug6-09, 04:49 PM
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Quote Quote by optics.tech View Post
Also, the trigonometry substitution method, but the denominator is not in [tex]\sqrt{x^2 + 1}[/tex] form.
Wait -- are you saying you tried it and failed? (If so, would you show your work, please?)

Or are you saying you didn't actually try it at all?

Hurkyl
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Aug6-09, 04:50 PM
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How to Integrate [1/(x^2 + 1)] dx?


Quote Quote by optics.tech View Post
But the x variable is still exist.
Don't you have an equation relating x to the variable you want to write everything in?
optics.tech
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#5
Aug6-09, 08:24 PM
P: 80
OK, as you asked for it:

FIRST

u = x^2 + 1, du/dx = 2x, du/2x = dx

[tex]\int\frac{1}{x^2+1} \ dx = \int\frac{1}{u} \ \frac{du}{2x}[/tex]

I can not do further because the x variable is still exist.

SECOND

[tex]x = tan \ \theta[/tex]
[tex]\frac{dx}{d\theta}=sec^2\theta[/tex]
[tex]dx=sec^2\theta \ d\theta[/tex]

[tex]x^2+1=(tan \ \theta)^2+1=tan^2\theta+1=sec \ \theta[/tex]

Then

[tex]\int\frac{1}{x^2+1} \ dx=\int\frac{1}{sec \ \theta} \ (sec^2\theta \ d\theta) = \int sec \ \theta \ d\theta= ln(tan \ \theta+sec \ \theta) \ + \ C=ln[x+(x^2+1)]+C=ln(x^2+x+1)+C[/tex]
jgens
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#6
Aug6-09, 08:37 PM
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If you applied the correct identity it might help . . .

[tex]tan^2(x) + 1 = sec^2(x)[/tex]

Your integral equation then becomes,

[tex]\int\frac{1}{x^2 + 1} \; dx = \int \, d\theta[/tex]
tiny-tim
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#7
Aug7-09, 01:59 AM
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Quote Quote by jgens View Post
If you applied the correct identity it might help . . .
he he
optics.tech, learn your trigonometric identities !
optics.tech
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#8
Aug7-09, 10:09 AM
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It's my mistake

[tex]arc \ tan \ x+C[/tex]
semc
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#9
Aug18-09, 12:25 AM
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Isnt the answer just inverse tan x +C ? Why do you need to use substitution?
jgens
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#10
Aug18-09, 12:27 AM
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If you don't recognize the anti-derivative at first sight, the substitution allows you to evaluate the integral without too much difficulty.
G Diddy
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#11
Nov20-09, 10:56 AM
P: 3
This is a standard contour integral. Convert x to z and locate the poles at +/- i.
vish_al210
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#12
Sep20-11, 04:00 AM
P: 88
hi, This is an old thread.. So if u'd like me to post it elsewhere, do let me know..
I need to calculate the theta using inverse-tan. But since micro-controllers do not provide much computational freedom, I was looking to solve it as the integral of 1/(1+x^2).
Other than the fact that, integral of 1/(1+x^2) is arctan(x). But since i'd like to know arctan(x), could someone please help me to find the intergral in terms of x (non-trigonometric).
tiny-tim
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#13
Sep20-11, 07:37 AM
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Quote Quote by vish_al210 View Post
integral of 1/(1+x^2) is arctan(x). But since i'd like to know arctan(x), could someone please help me to find the intergral in terms of x (non-trigonometric).
hi vish_al210!

(try using the X2 icon just above the Reply box )

you could try expanding 1/(1+x2) as 1 - x2 - , and then integrating
kanika2217
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#14
Nov8-11, 10:33 AM
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there is a direct formula for these kind of questins i.e integrate[1/(a^2+x^2)]dx = 1/a[arctan(x/a)]
so u can assume a=1.....so ur ans vil b (arctan x)......jst dis..
tiny-tim
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Nov8-11, 10:51 AM
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hi kanika2217! welcome to pf!
Quote Quote by kanika2217 View Post
there is a direct formula for these kind of questins
yes we know, but we're all trying to do it the way the ancient greeks would have done!

(btw, please don't use txt spelling on this forum it's against the forum rules )
kanika2217
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#16
Nov8-11, 10:54 AM
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hi!
v olso have a drct frmila for these kind of ques...i.e.integration[1/(a^2 + x^2)]dx = (1/a)(arctan (x/a)...so you can assume a=1 here and hence the ans wud b 'arctan x'....
M1991
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#17
Feb10-12, 05:40 AM
P: 1
Quote Quote by optics.tech View Post
Hi everyone,

Can you tell me how to integrate the following equation?

[tex]\int\frac{1}{x^2 + 1} \ dx[/tex]

I've tried the substitution method, u = x^2 + 1, du/dx = 2x. But the x variable is still exist.

Also, the trigonometry substitution method, but the denominator is not in [tex]\sqrt{x^2 + 1}[/tex] form.

Thanks in advance

Huygen
The answer is : Ln (x^2+1)/2x
tiny-tim
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#18
Feb10-12, 05:57 AM
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Hi M1991! Welcome to PF!

(try using the X2 button just above the Reply box )
Quote Quote by M1991 View Post
The answer is : Ln (x^2+1)/2x
no, that's ln(x2 + 1) - ln(2) - ln(x)
its derivative is 2x/(x2 + 1) - 1/x


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