What is the Integral of 1/{x-(1-x^2)0.5} and How Can It Be Simplified?

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Homework Help Overview

The discussion revolves around the integral of the function \( \frac{1}{x - \sqrt{1 - x^2}} \), exploring potential simplifications and methods for integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss simplifying the integral by multiplying by a conjugate and consider the possibility of using partial fractions. There is also mention of using trigonometric substitutions and the Weierstrass substitution for integration.

Discussion Status

Some participants have offered guidance on potential methods, such as trigonometric substitutions, while others question the validity of using partial fractions due to the nature of the numerator in the fraction. Multiple approaches are being explored without a clear consensus on the best method.

Contextual Notes

Participants note the complexity of the integration process and the challenges associated with it, including the length of steps involved in online resources.

haleycomet2
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Homework Statement


[itex]\int[/itex][itex]\frac{dx}{x-\sqrt{1-x^2}}[/itex]


Homework Equations



The Attempt at a Solution



i try to simplify it by multiplying [itex]\frac{x+\sqrt{1-x^2}}{x+\sqrt{1-x^2}}[/itex],becoming =[itex]\frac{x+\sqrt{1-x^2}}{2x^2-1}[/itex]
=[itex]\int[/itex]{[itex]\frac{x}{2x^2-1}[/itex]+[itex]\frac{\sqrt{1-x^2}}{2x^2-1}[/itex]}dx

ps:
a.Can i partial fraction the last term to [itex]\frac{A}{x-\frac{1}{\sqrt{2}}}[/itex]+[itex]\frac{B}{x+\frac{1}{\sqrt{2}}}[/itex]??
b.i try to integrate by using wolfram alpha online ,but the steps is incredibly long..
does it exist any other simpler way??
 

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I checked W-A, and they did it the same way I would have: Begin with a trig substitution like x=sin(u), then use the Weierstrass substitution: v=tan(u/2). It will be long and difficult, but it's possible.
 
haleycomet2 said:
a.Can i partial fraction the last term to [itex]\frac{A}{x-\frac{1}{\sqrt{2}}}[/itex]+[itex]\frac{B}{x+\frac{1}{\sqrt{2}}}[/itex]??
No, because both the numerator and denominator of the original fraction must be polynomials, and in
[tex]\frac{\sqrt{1-x^2}}{2x^2-1}[/tex]
the numerator is not a polynomial.
 
o,i see.Thanks a lot:smile:
 

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