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

by optics.tech
Tags: integration
 P: 80 Hi everyone, Can you tell me how to integrate the following equation? $$\int\frac{1}{x^2 + 1} \ dx$$ 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 $$\sqrt{x^2 + 1}$$ form. Thanks in advance Huygen
 Sci Advisor HW Helper Thanks P: 26,157 Hi Huygen! Try x = tanu.
Emeritus
PF Gold
P: 16,092
 Quote by optics.tech Also, the trigonometry substitution method, but the denominator is not in $$\sqrt{x^2 + 1}$$ 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?

Emeritus
PF Gold
P: 16,092
How to Integrate [1/(x^2 + 1)] dx?

 Quote by optics.tech But the x variable is still exist.
Don't you have an equation relating x to the variable you want to write everything in?
 P: 80 OK, as you asked for it: FIRST u = x^2 + 1, du/dx = 2x, du/2x = dx $$\int\frac{1}{x^2+1} \ dx = \int\frac{1}{u} \ \frac{du}{2x}$$ I can not do further because the x variable is still exist. SECOND $$x = tan \ \theta$$ $$\frac{dx}{d\theta}=sec^2\theta$$ $$dx=sec^2\theta \ d\theta$$ $$x^2+1=(tan \ \theta)^2+1=tan^2\theta+1=sec \ \theta$$ Then $$\int\frac{1}{x^2+1} \ dx=\int\frac{1}{sec \ \theta} \ (sec^2\theta \ d\theta) = \int sec \ \theta \ d\theta= ln(tan \ \theta+sec \ \theta) \ + \ C=ln[x+(x^2+1)]+C=ln(x^2+x+1)+C$$
 P: 1,622 If you applied the correct identity it might help . . . $$tan^2(x) + 1 = sec^2(x)$$ Your integral equation then becomes, $$\int\frac{1}{x^2 + 1} \; dx = \int \, d\theta$$
HW Helper
Thanks
P: 26,157
 Quote by jgens If you applied the correct identity it might help . . .
he he
optics.tech, learn your trigonometric identities !
 P: 80 It's my mistake $$arc \ tan \ x+C$$
 P: 329 Isnt the answer just inverse tan x +C ? Why do you need to use substitution?
 P: 1,622 If you don't recognize the anti-derivative at first sight, the substitution allows you to evaluate the integral without too much difficulty.
 P: 3 This is a standard contour integral. Convert x to z and locate the poles at +/- i.
 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).
HW Helper
Thanks
P: 26,157
 Quote by vish_al210 … 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
 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..
HW Helper
Thanks
P: 26,157
hi kanika2217! welcome to pf!
 Quote by kanika2217 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 )
 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'....
P: 1
 Quote by optics.tech Hi everyone, Can you tell me how to integrate the following equation? $$\int\frac{1}{x^2 + 1} \ dx$$ 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 $$\sqrt{x^2 + 1}$$ form. Thanks in advance Huygen
The answer is : Ln (x^2+1)/2x