How to Integrate [1/(x^2 + 1)] dx? (2024)

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Related to How to Integrate [1/(x^2 + 1)] dx? 1. What is the Integral of \(\frac{1}{x^2 + 1}\)? 2. Why is \(\arctan(x)\) the Antiderivative of \(\frac{1}{x^2 + 1}\)? 3. How Do You Approach This Integral Using Substitution? 4. Can This Integral Be Solved Using Trigonometric Identities? 5. Is It Necessary to Know the Bounds of Integration? 6. What are Common Mistakes to Avoid When Integrating \(\frac{1}{x^2 + 1}\)? Similar threads References
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    DxIntegrateIntegrationSubstitution method

In summary, the conversation discusses different methods for integrating the equation \int\frac{1}{x^2 + 1} \ dx, including substitution and trigonometric substitution. The direct formula \frac{1}{x^2+a^2}= \frac{1}{a} * \arctan(\frac{x}{a}) is suggested as a possible solution.

  • #1

optics.tech

79
1

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

  • #2

tiny-tim

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

Try x = tanu. How to Integrate [1/(x^2 + 1)] dx? (2)​

  • #3

Hurkyl

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optics.tech said:

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?

  • #4

Hurkyl

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optics.tech said:

But the x variable is still exist.

Don't you have an equation relating x to the variable you want to write everything in?

  • #5

optics.tech

79
1

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

jgens

Gold Member

1,593
50

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

tiny-tim

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jgens said:

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

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

optics.tech, learn your trigonometric identities ! How to Integrate [1/(x^2 + 1)] dx? (4)​

  • #8

optics.tech

79
1

It's my mistake How to Integrate [1/(x^2 + 1)] dx? (5)

[tex]arc \ tan \ x+C[/tex]

  • #9

semc

368
5

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

  • #10

jgens

Gold Member

1,593
50

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
0

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

  • #12

vish_al210

85
0

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

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vish_al210 said:

… 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! How to Integrate [1/(x^2 + 1)] dx? (6)

(try using the X2 icon just above the Reply box How to Integrate [1/(x^2 + 1)] dx? (7))

you could try expanding 1/(1+x2) as 1 - x2 - … , and then integrating How to Integrate [1/(x^2 + 1)] dx? (8)

  • #14

kanika2217

3
0

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

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welcome to pf!

hi kanika2217! welcome to pf! How to Integrate [1/(x^2 + 1)] dx? (9)

kanika2217 said:

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! How to Integrate [1/(x^2 + 1)] dx? (10)

(btw, please don't use txt spelling on this forum … it's against the forum rules How to Integrate [1/(x^2 + 1)] dx? (11))

  • #16

kanika2217

3
0

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
0

optics.tech said:

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

  • #18

tiny-tim

Science Advisor

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258

Welcome to PF!

Hi M1991! Welcome to PF! How to Integrate [1/(x^2 + 1)] dx? (12)

(try using the X2 button just above the Reply box How to Integrate [1/(x^2 + 1)] dx? (13))

M1991 said:

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 How to Integrate [1/(x^2 + 1)] dx? (14)​

  • #19

torpedala

1
0

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

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

1. What is the Integral of \(\frac{1}{x^2 + 1}\)?

The integral of \(\frac{1}{x^2 + 1}\) with respect to \(x\) is \(\arctan(x) + C\), where \(C\) is the constant of integration. This integral is a standard result in calculus, derived from the derivative of the arctan (or inverse tangent) function.

2. Why is \(\arctan(x)\) the Antiderivative of \(\frac{1}{x^2 + 1}\)?

\(\arctan(x)\) is the antiderivative of \(\frac{1}{x^2 + 1}\) because the derivative of \(\arctan(x)\) is \(\frac{1}{x^2 + 1}\). This relationship is a fundamental part of trigonometric integration, reflecting how the slopes of the tangent function relate to the values of the arctan function.

3. How Do You Approach This Integral Using Substitution?

For the integral \(\frac{1}{x^2 + 1}\) dx, a substitution is not necessary as it directly matches the standard form. However, if the integral were modified (e.g., \(\frac{1}{(ax^2 + 1)}\) dx), substitution methods might be needed to bring it into a form where the arctan function can be easily recognized.

4. Can This Integral Be Solved Using Trigonometric Identities?

Trigonometric identities aren't typically used to solve \(\frac{1}{x^2 + 1}\) dx directly, as the integral already matches the derivative of the arctan function. However, trigonometric identities might be helpful in more complex integrals involving expressions similar to \(x^2 + 1\) in the denominator.

5. Is It Necessary to Know the Bounds of Integration?

If you are calculating a definite integral, then yes, you need to know the bounds of integration. The integral \(\int_a^b \frac{1}{x^2 + 1} dx\) will give the area under the curve between \(x = a\) and \(x = b\). For an indefinite integral, bounds are not required, and the result will include the constant of integration \(C\).

6. What are Common Mistakes to Avoid When Integrating \(\frac{1}{x^2 + 1}\)?

Common mistakes include overcomplicating the integral by attempting unnecessary substitutions or transformations. Remembering that \(\frac{1}{x^2 + 1}\) is the derivative of \(\arctan(x)\) can simplify the integration process significantly.

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                      How to Integrate [1/(x^2 + 1)] dx? (2024)

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