Trigonomic Integration question

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Homework Help Overview

The discussion revolves around the integral of the function ((1+x)/(1+(x^2))) with respect to x, which falls under the subject area of trigonometric integration.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the possibility of splitting the integral into two separate integrals for easier handling. Some express uncertainty about how to proceed with the integration, while others suggest specific substitutions and methods for each part of the integral.

Discussion Status

There are multiple interpretations being explored regarding how to approach the integral. Some participants have provided guidance on splitting the integral and suggested methods for integration, but there is no explicit consensus on the best approach yet.

Contextual Notes

The original poster expresses a need for assistance due to uncertainty about how to start the problem, indicating a potential lack of familiarity with the topic.

jordan123
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Homework Statement


Sorry for the lack of latex.

The question is this.

Integral of ((1+x)/(1+(x^2))) dx

If someone could put that together on latex, that would be awesome!


The Attempt at a Solution


Im not exactly sure what to do, at all. So really any help is great. Even just a hint and I will probably get it. Thanks!
 
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Split into two integrals.
[tex]\int \frac{1 + x}{1 + x^2}dx[/tex]
[tex]= \int \frac{1}{1 + x^2} dx + \int \frac{x}{1 + x^2} dx[/tex]

The first is pretty straightforward. The second requires only an ordinary substitution.
 
((1+x)/(1+(x^2))) dx = (1/(1+(x^2)) dx+(x)/(1+(x^2)) dx

Yes, first part is trigonometric and second is log..
 
Do you mean

[tex] <br /> \int \frac{1 + x} {1 + x^2} dx<br /> [/tex]

Re-write the integrand as two terms
 
Mark44 said:
Split into two integrals.
[tex]\int \frac{1 + x}{1 + x^2}dx[/tex]
[tex]= \int \frac{1}{1 + x^2} dx + \int \frac{x}{1 + x^2} dx[/tex]

The first is pretty straightforward. The second requires only an ordinary substitution.

lol, thanks.
 

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