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How to Integrate [1/(x^2 + 1)] dx?by optics.tech
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#1
Aug609, 02:45 PM

P: 80

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 


#3
Aug609, 04:49 PM

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PF Gold
P: 16,092

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


#4
Aug609, 04:50 PM

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PF Gold
P: 16,092

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



#5
Aug609, 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] 


#6
Aug609, 08:37 PM

P: 1,622

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] 


#7
Aug709, 01:59 AM

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P: 26,157

optics.tech, learn your trigonometric identities ! 


#8
Aug709, 10:09 AM

P: 80

It's my mistake
[tex]arc \ tan \ x+C[/tex] 


#9
Aug1809, 12:25 AM

P: 329

Isnt the answer just inverse tan x +C ? Why do you need to use substitution?



#10
Aug1809, 12:27 AM

P: 1,622

If you don't recognize the antiderivative at first sight, the substitution allows you to evaluate the integral without too much difficulty.



#11
Nov2009, 10:56 AM

P: 3

This is a standard contour integral. Convert x to z and locate the poles at +/ i.



#12
Sep2011, 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 inversetan. But since microcontrollers 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 (nontrigonometric). 


#13
Sep2011, 07:37 AM

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(try using the X^{2} icon just above the Reply box ) you could try expanding 1/(1+x^{2}) as 1  x^{2}  … , and then integrating 


#14
Nov811, 10:33 AM

P: 3

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.. 


#15
Nov811, 10:51 AM

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P: 26,157

hi kanika2217! welcome to pf!
(btw, please don't use txt spelling on this forum … it's against the forum rules ) 


#16
Nov811, 10:54 AM

P: 3

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'.... 


#17
Feb1012, 05:40 AM

P: 1




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