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

[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

 

[tiny-tim]

Hi Huygen!

Try x = tanu. ​

 

[Hurkyl]

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?

 

[Hurkyl]

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]

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

 

[jgens]

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

 

[tiny-tim]

If you applied the correct identity it might help . . .

he he

optics.tech, learn your trigonometric identities ! ​

 

[optics.tech]

It's my mistake

arc \ tan \ x+C

 

[semc]

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

 

[jgens]

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]

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

 

[vish_al210]

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]

… 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]

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]

welcome to pf!

hi kanika2217! welcome to pf!

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]

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]

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

 

[tiny-tim]

Welcome to PF!

Hi M1991! Welcome to PF!

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

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 ​

 

[torpedala]

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