How do I do this integration using u sub?

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Homework Help Overview

The discussion revolves around the integration of the function ∫x^2√(2+x) using substitution methods. Participants are exploring the use of u-substitution to simplify the integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to identify a suitable substitution for the integral but expresses confusion over the results of their substitutions. Some participants question the clarity of the original poster's reasoning and request further explanation of their steps.

Discussion Status

There is an ongoing exploration of different substitution methods. Some participants have suggested specific substitutions, while others are seeking clarification on the original poster's attempts. No consensus has been reached, but there are indications of productive dialogue regarding the approach to the problem.

Contextual Notes

Participants are discussing the implications of different substitutions and their derivatives, indicating a focus on understanding the mechanics of u-substitution in the context of integration.

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Homework Statement


∫x^2√(2+x)

using u sub


Homework Equations


∫x^2√(2+x)


The Attempt at a Solution



I can't seem to find anything to use for a u sub.

if I sub 2+x I just get 1, and if I sub x^2 I just get 2x

If I do √(2+x) I just get 1/2(1/√(2+x))
 
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Frankly, I really don't know what you mean by
"if I sub 2+x I just get 1, and if I sub x^2 I just get 2x
If I do √(2+x) I just get 1/2(1/√(2+x)) "

You get 1 for what? The integral? Show your work! HOW did you get "1"?
 
Let [itex]J:=\int x^2\sqrt{x+1}dx[/itex]. Put [itex]u=x+1\Rightarrow du=dx[/itex] to get [itex]J=\int (u-1)^2\sqrt{u}du=\int \left( u^\frac{5}{2}-2u^\frac{3}{2}+u^\frac{1}{2}\right) du[/itex]
 
HallsofIvy said:
Frankly, I really don't know what you mean by
"if I sub 2+x I just get 1, and if I sub x^2 I just get 2x
If I do √(2+x) I just get 1/2(1/√(2+x)) "

You get 1 for what? The integral? Show your work! HOW did you get "1"?

The derivative of 2+x = 1
 
benorin said:
Let [itex]J:=\int x^2\sqrt{x+1}dx[/itex]. Put [itex]u=x+1\Rightarrow du=dx[/itex] to get [itex]J=\int (u-1)^2\sqrt{u}du=\int \left( u^\frac{5}{2}-2u^\frac{3}{2}+u^\frac{1}{2}\right)[/itex]

Thanks, so I use change of variable to get my answers, I should've seen that.
 

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