Antiderivative Help: Calculate Integral

In summary: 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.
  • #1
OmniNewton
105
5

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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  • #2
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.
 
  • #3
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.
 
  • #4
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]
 
  • #5
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
 
  • #6
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.
 
  • #7
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..
 

1. What is an antiderivative?

An antiderivative, also known as an indefinite integral, is the reverse process of taking a derivative. It is a function that, when differentiated, gives the original function.

2. How do you calculate an antiderivative?

To calculate an antiderivative, you can use the power rule, product rule, quotient rule, and chain rule. You can also use integration techniques such as substitution, integration by parts, and partial fractions.

3. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration, while an indefinite integral does not. A definite integral gives a numerical value, whereas an indefinite integral gives a function.

4. Can all functions have an antiderivative?

No, not all functions have an antiderivative. For a function to have an antiderivative, it must be continuous and differentiable on its entire domain.

5. How can calculating an antiderivative be useful?

Calculating an antiderivative can be useful in finding the area under a curve, determining the velocity and acceleration of an object, and solving differential equations in science and engineering applications.

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