Integrating: Solving the Problem of Ln(u) in Answer

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Homework Help Overview

The discussion revolves around the integration of the function \(\int(\frac{x}{\sqrt{1-x^{2}}})dx\), focusing on the challenges faced in simplifying the expression and addressing the presence of \(\ln(u)\) in the final answer.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use substitution with \(u = \sqrt{1-x^2}\) but encounters difficulties with the resulting logarithmic term. Other participants suggest alternative substitutions and transformations to simplify the integral.

Discussion Status

Participants are exploring different substitution methods and transformations to address the integration problem. Some guidance has been offered regarding alternative approaches, but there is no explicit consensus on the best method yet.

Contextual Notes

There are indications of confusion regarding the integration process and the presence of logarithmic terms, which may reflect underlying assumptions about the integration techniques being applied.

crm08
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Homework Statement



[tex]\int(\frac{x}{\sqrt{1-x^{2}}})dx[/tex]

Homework Equations





The Attempt at a Solution



My calculator tells me that the answer should be -sqrt(1-x^2) but if I pick u = sqrt(1-x^2), then dx = (sqrt(1-x^2)*du)/x, which leaves me with -integral((sqrt(1-x^2)/u)du), the problem I am having is getting rid of the "ln(u)" in my final answer, any suggestions?
 
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If u^2=1-x^2
2u du =-2x dx => - u du = x dx

Now you'd just get

[tex]\frac{-u}{u} du[/tex]
 
ok got it, thank you
 
rock.freak667 said:
If u^2=1-x^2
2u du =-2x dx => - u du = x dx

Now you'd just get

[tex]\frac{-u}{u} du[/tex]
Another substitution that works is u = 1 - x^2, du = -2xdx.
The integrand then becomes -(1/2)du/u^(1/2), which is also an easy one to integrate.
 

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