How Do You Integrate \( \frac{x}{(1-x^2)} \) dx?

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Homework Help Overview

The discussion revolves around the integration of the function \( \frac{x}{(1-x^2)} \) with respect to \( x \). Participants explore various integration techniques and substitutions to approach the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss integration by parts and substitution methods. One participant attempts integration by parts but struggles with defining the function \( b(x) \). Another suggests a substitution \( y = x^2 \) to simplify the integral, while a third proposes letting \( u = 1 - x^2 \) for a different approach.

Discussion Status

The discussion is active, with participants sharing different substitution methods and clarifying each other's approaches. There is acknowledgment of confusion regarding the initial steps, but no consensus has been reached on a single method.

Contextual Notes

Participants express uncertainty about the integration process and the forms of substitutions to use, indicating a need for further exploration of these methods.

tomwilliam2
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Homework Statement


$$\int \frac{x}{(1-x^2)} dx$$

Homework Equations



Integration by parts, by substitution, etc.

The Attempt at a Solution



I just can't remember how to begin this integration. I tried doing integration by parts, where
$$a(x) = x$$
$$a'(x) = 1$$
$$b(x) = $$
$$b'(x) = \frac{1}{(1-x^2)}$$
$$\int \frac{x}{(1-x^2)}dx = xb(x) - \int \frac{1}{(1-x^2)} dx$$

But couldn't work out what form b(x) should take.
Thanks in advance for any help
 
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If you have seen enough of these integrals, they can essentially done "by observation"

If not, substitution is a good method to use. Substitute [itex]y = x^{2}[/itex]. Then [itex]dy = 2x dx[/itex]
Then the integral becomes:
[tex]\int \frac{1}{2}\frac{1}{1-y}dy[/tex]

Does that now look more doable?
 
Last edited:
Why not let u= 1- x^2 itself? Then du= -2xdx so that xdx= -(1/2)du do that
[itex]-\frac{1}{2}\int \frac{du}{u}[/itex]
 
Ah yes, thanks...that makes sense. I wasn't getting where the sqrt cam from in the first response.
Thanks to both of you
 
tomwilliam2 said:
I wasn't getting where the sqrt cam from in the first response
Oops, typo, my bad.
 

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