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

1. ### optics.tech

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.

Huygen

2. ### tiny-tim

26,041
Hi Huygen!

Try x = tanu.

3. ### Hurkyl

16,089
Staff Emeritus
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?

4. ### Hurkyl

16,089
Staff Emeritus
Don't you have an equation relating x to the variable you want to write everything in?

5. ### optics.tech

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

$$\int\frac{1}{x^2+1} \ dx=\int\frac{1}{sec \ \theta} \ (sec^2\theta \ d\theta) = \int sec \ \theta \ d\theta= ln(tan \ \theta+sec \ \theta) \ + \ C=ln[x+(x^2+1)]+C=ln(x^2+x+1)+C$$

civileng likes this.
6. ### jgens

1,621
If you applied the correct identity it might help . . .

$$tan^2(x) + 1 = sec^2(x)$$

$$\int\frac{1}{x^2 + 1} \; dx = \int \, d\theta$$

civileng likes this.
7. ### tiny-tim

26,041
he he

optics.tech, learn your trigonometric identities !

8. ### optics.tech

80
It's my mistake

$$arc \ tan \ x+C$$

9. ### semc

334
Isnt the answer just inverse tan x +C ? Why do you need to use substitution?

10. ### jgens

1,621
If you don't recognize the anti-derivative at first sight, the substitution allows you to evaluate the integral without too much difficulty.

11. ### G Diddy

3
This is a standard contour integral. Convert x to z and locate the poles at +/- i.

12. ### vish_al210

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).

13. ### tiny-tim

26,041
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

14. ### kanika2217

3
Re: How to Integrate[1/(x^2 + 1)] dx?

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. ### tiny-tim

26,041
welcome to pf!

hi kanika2217! welcome to pf!
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 )

16. ### kanika2217

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. ### M1991

1
The answer is : Ln (x^2+1)/2x

18. ### tiny-tim

26,041
Welcome to PF!

Hi M1991! Welcome to PF!

(try using the X2 button just above the Reply box )
no, that's ln(x2 + 1) - ln(2) - ln(x) …

its derivative is 2x/(x2 + 1) - 1/x

19. ### torpedala

1
Direct formula: 1/(x2+a2)=1/a * arctg( x/a ) to Integrate [1/(x^2 + 1)] dx

Last edited by a moderator: Feb 10, 2015