Antiderivative and Indefinite Integration

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Homework Help Overview

The discussion revolves around the integration of a rational function involving a square root in the denominator, specifically the integral ∫(x³ - 3x² + x + 1)/√x dx. Participants are exploring the process of finding the antiderivative and addressing the nuances of indefinite integration.

Discussion Character

  • Mathematical reasoning, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to simplify the integral by rewriting terms and integrating them separately. There are questions regarding the correct application of integration rules, particularly in relation to the powers of x and the constants involved in the integration process.

Discussion Status

Some participants are providing feedback on the integration steps taken by others, pointing out potential errors in the application of integration techniques. There is an ongoing exploration of the correct interpretation of the integrals involved, with no clear consensus reached yet.

Contextual Notes

Participants are navigating through the rules of integration, including the handling of powers and constants. There is some confusion regarding the rewriting of integrals and the implications of different forms of the integrals presented.

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Homework Statement



∫(x3 - 3x2 + x + 1)/√x dx

The Attempt at a Solution



∫x3-1/2 - 3∫x2-1/2 + ∫x1-1/2 + ∫x1-1/2

∫x5/2 - 3∫x3/2 + ∫x1/2 + ∫1/2

(x7/2)7/2 - (3x5/2)5/2 + (x3/2)3/2 + (x3/2)3/2 + C

(2x7/2)/7 - (6x5/2)/5 + 2x3/2)/3 + (2x3/2)/3 + C

Thank you so much!
 
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First, the last integral you wrote, should be ∫{\frac{1}{√x}}
Second, When integrating, you divide by the new power, you are multiplying, so you should 2/7, 2/5.. etc..

Edit: I am not sure what you did on the line just after integrating, but the line after it is mostly right, except for that last part.
 
I don't know if I understand it.

If I have:

∫x3/√x dx

= ∫x3 - 1/2
= ∫x5/2
= (x5/2 + 1)/ 5/2 +1
= (x7/2)/7/2 + C

Simplyfying:

2x7/2/7 + C

Is this correct?

If not, what am I doing wrong?

Also, I am not sure how to proceed with ∫1/√x dx

If I have ∫1 dx:

= 1 + C

So for ∫1/√x dx I'll have:

=x/√x
=x . x- 1/2
=x1/2
=(x1/2+ 1)/1/2 +1 + C
 
The first part was right. for ∫1/√x dx. rewrite it as ∫x-1/2dx

EDIT: Nowhere in the equation would you have ∫1dx, but if you did. it would be x+c, not 1+c.
 

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