Simple Integral Solution: Finding the Value of a in ∫1/(x^2+a)dx Formula

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Homework Help Overview

The discussion revolves around evaluating the integral of the function \(\int \frac{1}{x^2+a} dx\), focusing on the parameter \(a\) and its implications on the integral's form. The subject area is calculus, specifically integral calculus.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the conversion of the integral into a specific form involving the square root of \(a\) and question the relevance of choosing the positive or negative root. There is a discussion about whether the choice affects the outcome due to the properties of the arctangent function.

Discussion Status

Some participants have provided guidance on the implications of the odd function property of the arctangent, suggesting that the choice of root may not matter in the final expression. However, there is no explicit consensus on the necessity of this consideration.

Contextual Notes

Participants are navigating the implications of the integral's form and the nature of the parameter \(a\). The discussion reflects an exploration of assumptions regarding the function's behavior and the mathematical properties involved.

Mr Davis 97
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Homework Statement


##\displaystyle \int \frac{1}{x^2+a} dx##

Homework Equations

The Attempt at a Solution


I know that I can convert this to the form ##\displaystyle \int \frac{1}{x^2+(\pm \sqrt{a})^2} dx## = ##\displaystyle \frac{1}{\pm \sqrt{a}} \arctan (\frac{x}{\pm \sqrt{a}}) + C##, but I don't know whether to take the positive or the negative root.
 
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Ask yourself whether it matters or not.
 
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Orodruin said:
Ask yourself whether it matters or not.
Does it not matter since ##\arctan## is an odd function, so in either case you get ##\displaystyle \frac{1}{\sqrt{a}} \arctan (\frac{x}{\sqrt{a}}) + C##?
 
Indeed.
 
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