Solve Indefinite Integral: $\int\frac{\arctan x}{1+x^2}dx$

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Discussion Overview

The discussion revolves around solving the indefinite integral $\int\frac{\arctan x}{1+x^2}dx$. Participants explore various substitution methods and notation for expressing the result.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant inquires about how to approach the integral, indicating initial attempts at substitution were unsuccessful.
  • Another participant suggests using the substitution $u = \tan^{-1}(x)$ and prompts the first participant to consider the derivative of this function.
  • A subsequent participant confirms understanding of the substitution but seeks clarification on its application.
  • Following the substitution, a participant notes that the integral simplifies to $\int u \, du$ and encourages others to continue from there.
  • One participant presents their solution as $\frac{1}{2}(\tan^{-1}x)^2 + C$.
  • Another participant provides a similar answer but uses a different notation, $\frac{1}{2}(Tan^{-1})^2 + C$, and comments on the clarity of notation, suggesting $\frac{1}{2}(\arctan^{2}(x)) + C$ as preferable.
  • A participant reiterates the preference for clearer notation and mentions that "atn(x)" is also a suitable alternative.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the integral using substitution, but there is some disagreement regarding the notation used to express the final answer.

Contextual Notes

There are unresolved aspects regarding the clarity of notation and the initial approach to substitution, as some participants express confusion about the application of the method.

paulmdrdo1
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how would i go about solving this

$\displaystyle\int\frac{\arctan x}{1+x^2}dx$?

i tried substitution but i didn't work.
 
Last edited:
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That's very strange. Did you try the obvious substitution u= tan^{-1}(x). What is the derivative of tan^{1}(x)? (And do you understand that tan^{-1}(x) here is the arctangent, NOT the \frac{1}{tan(x)}?!)
 
yes, i know that. but how do i use substitution here?
 
Following HoI's suggestion, we get that $u=\tan^{-1}(x)$, and
$$du= \frac{1}{1+x^{2}} \, dx,$$
and the integral becomes
$$\int u \, du.$$
Can you continue?
 
this is my answer,

$\displaystyle\frac{1}{2}(\tan^{-1}x)^2+C$
 
Last edited:
paulmdrdo said:
this is my answer,

$\displaystyle\frac{1}{2}(Tan^{-1})^2+C$

That is the correct idea, but awful notation: $\displaystyle\frac{1}{2}(\arctan^{2}(x))+C$
 
Plato said:
That is the correct idea, but awful notation: $\displaystyle\frac{1}{2}(\arctan^{2}(x))+C$
I find that "atn(x)" works quite well also.

-Dan
 

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