Antiderivative Help: Calculate Integral

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Homework Help Overview

The discussion revolves around calculating an integral, with participants exploring various methods of integration, including u-substitution and factoring. The problem appears to involve a quadratic expression under a root sign.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss attempts at u-substitution, with one suggesting letting u = -3 + 4/x - 1/x^2. Others propose factoring the quadratic and completing the square as potential strategies. There is uncertainty about the effectiveness of these approaches.

Discussion Status

The conversation is ongoing, with participants sharing their attempts and considering different methods. Some guidance has been offered regarding factoring and substitutions, but no consensus has been reached on the best approach.

Contextual Notes

Participants express confusion about the direction of their attempts and the complexity of the problem, indicating that they may be working under constraints typical of homework assignments.

OmniNewton
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Homework Statement


Calculate the following integral:
b1719ffb92379d42d595382b30430c6f.png


Homework Equations


N/A

The Attempt at a Solution


b1719ffb92379d42d595382b30430c6f.png

728c33c9e6b78570668463ea8929d8f2.png

36a8f99e810595ddc5eef07ea876eefe.png

from this point I tried a u-substitution by letting u = -3 + 4/x -1/x^2 but this seemed to fail.

Are any suggestions possible?
 
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OmniNewton said:

Homework Statement


Calculate the following integral:
b1719ffb92379d42d595382b30430c6f.png


Homework Equations


N/A

The Attempt at a Solution


b1719ffb92379d42d595382b30430c6f.png

728c33c9e6b78570668463ea8929d8f2.png

36a8f99e810595ddc5eef07ea876eefe.png

from this point I tried a u-substitution by letting u = -3 + 4/x -1/x^2 but this seemed to fail.

Are any suggestions possible?
Try factoring the quadratic under the root sign first. You should be able to find two linear factors.
 
SteamKing said:
Try factoring the quadratic under the root sign first. You should be able to find two linear factors.
Alright thank you will attempt that now.
 
OmniNewton said:
Alright thank you will attempt that now.
Alright I have the indefinite integral factored for the argument however I am unsure how to proceed.
360ea395d27c22bc3e3818285928f822.png

I'm going to let u = x-1 which means x = u + 1 and for the second factor (3x-1) = (3x - 1 - 2 + 2) = [3(x-1) + 2]
 
OmniNewton said:
Alright I have the indefinite integral factored for the argument however I am unsure how to proceed.
360ea395d27c22bc3e3818285928f822.png

I'm going to let u = x-1 which means x = u + 1 and for the second factor (3x-1) = (3x - 1 - 2 + 2) = [3(x-1) + 2]

I may not be approaching this the right direction since after attempting this method I seem to be unable to proceed
 
There's usually several substitutions that work for these types of problems. The simplest one seems to be ##t=1/x##. At least that seem to work for me (but I did this rather quickly so I may have done something wrong).

Another approach that should work would be is to complete the square and then use the standard substitution ##\arcsin## but in this case it seems to be rather complicated and you would need another substitution ##s=\tan t/2## after as well.
 
That looks like I would complete the square of the term under the square root to get something to match arcsec or arccsc. 1/(u √ (u2-a2) is the one I believe i am thinking of..
 

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