How Can I Solve ∫[6/(x^2+3)^2] Using Substitution?

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Homework Help Overview

The discussion revolves around solving the integral ∫[6/(x^2+3)^2] using substitution techniques. The subject area includes calculus, specifically integral calculus and trigonometric substitution.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different substitution methods, with one suggesting x = 3tan(u) and another proposing x = √3tan(θ). There is a discussion about the effectiveness of these substitutions and the complexity of the resulting integrals.

Discussion Status

The discussion is active, with participants sharing their approaches and questioning the validity of different substitution methods. Some guidance is offered regarding the potential complications of certain substitutions, but no consensus on a single method has been reached.

Contextual Notes

Participants express concern over the effectiveness of their chosen substitutions and the resulting complexity of the integrals they are attempting to evaluate. There is an acknowledgment that different substitutions may lead to varying levels of difficulty in solving the integral.

badtwistoffate
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in int[6/(x^2+3)^2] i use 3tanU=x, dx=3sec(u)^2 du
using that i get to ... int[1+cos2u]= u + sin2u/2 = u +sin u cos u... but subing back in with trig doesn't give me the right answer... any help with this method?
 
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[tex]I = \int {\frac{6}{{\left( {x^2 + 3} \right)^2 }}dx}[/tex]

Instead of setting x = 3tan(u) try...

[tex] x = \sqrt 3 \tan \theta \Rightarrow dx = \sqrt 3 \sec ^2 \theta d\theta [/tex]
 
darn it, after looking at my work i was afraid someone would say that... as i didnt relize that would be better until i finished it and noticed it was coming out right... does that mean you can't do it my way:-/
 
There's really no reason why it can't be done your way. It's just that some substitutions will require you to evaluate much more complicated integrals than if you were to use a substitution which obviously simplifies the integrand.
 

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