Integration of x^2 dx / (1-x)^1/2

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Homework Help Overview

The discussion revolves around the integration of the function x^2 dx / (1-x)^(1/2), focusing on the application of u-substitution and the subsequent steps in the integration process.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the use of u-substitution, with varying interpretations of how to express dx in terms of du. Questions arise regarding the correct manipulation of the integral and the handling of the numerator x^2.

Discussion Status

Participants have provided hints and suggestions for reworking the substitution and integration steps. There is an ongoing exploration of the differentiation of results to verify correctness, with some expressing confusion over the outcomes of their calculations.

Contextual Notes

Some participants note discrepancies in their results compared to online calculators, indicating a potential misunderstanding in the substitution or integration process. There is also mention of the need to ensure proper handling of the expressions derived from the substitution.

ganondorf29
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Homework Statement



Integrate: x^2 dx / (1-x)^1/2

Homework Equations



U substitution

The Attempt at a Solution



First I defined u = (1-x)^1/2
du = -dx/2(1-x)^1/2
dx = -2(1-x)^1/2 du

then for x
u^2 = 1-x
x = 1-u^2

integral of: (1-u^2)^2/u
= (u^4-2u^2+1) / u
= u^3-2u+1/u

integrating:

u^4/4 - u^2 + 1/u

= ((1-x)^2)/4 - (1-x) + ln|1-x| + CIs this right?
 
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You can try differentiating your answer to check if it is correct..
 
I differentiated and the answer I got was wrong. Does anyone know what I did wrong?
 
looks like you should have gotten ,

ln| (1-x)^1/2 |

EDIT: I was wrong here, look at Dick's hint later. It seems pretty easy after his hint.
 
Last edited:
You figured out that dx=-2u*du. But then you never used it. There's no logs in the problem at all.
 
If dx = -2u*du that means that I would have to set u = (1-x)^1/2 and square both sides and then take the derivative, right? Well, if I do that, what do I do for the x^2 on the numerator?
 
ganondorf29 said:
If dx = -2u*du that means that I would have to set u = (1-x)^1/2 and square both sides and then take the derivative, right? Well, if I do that, what do I do for the x^2 on the numerator?

What?? No, you are doing everything right, but you forgot to replace the dx with an expression for du. You said "dx = -2(1-x)^1/2 du". Use that expression.
 
Ok, I reworked the dx part and I got

Integral of:
((1-u^2)^2*-2u*du)/u
= -2(1-u^2)^2
= -2(u^4-2u^2+1)

= -(2/5)(u^5) + (4/3)(u^3) - 2u + C

= -(2/5)((1-x)^5) + (4/3)((1-x)^3) - 2(1-x) + C

But when I differentiate this I am not getting x^2/sqrt(1-x)
:confused:
 
u=sqrt(1-x). You substituted u=(1-x).
 
  • #10
After fixing the substitution

-(2/5)((sqrt(1-x))^5) + (4/3)((sqrt(1-x))^3) - 2(sqrt(1-x)) + C

d/dx of -(2/5)((sqrt(1-x))^5) + (4/3)((sqrt(1-x))^3) - 2(sqrt(1-x)) + C

= 1/sqrt(1-x) - (x+1)*(sqrt(1-x))

which isn't x^2/(sqrt(1-x))


I went to mathematica's online integral calculator and they gave this:
-(2/15)*(sqrt(1-x)*(x(3x+4)+8

My Ti-89 gave the same answer once factored. I still don't know what I'm doing wrong.
 
  • #11
1/sqrt(1-x) - (x+1)*(sqrt(1-x)). Multiply the numerator and denominator of the second term by sqrt(1-x). 1/sqrt(1-x)-(x+1)(1-x)/sqrt(1-x)=(1-(1-x^2))/sqrt(1-x)=x^2/sqrt(1-x). It's right. You've got too many machines telling you what to do.
 
  • #12
Thank you so much
 

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