Integrating 1/(x+x^3+2): Simplified Steps & Techniques

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In summary, the conversation discusses methods for integrating the given function, with one participant attempting to use partial fraction decomposition but finding it complicated. They ask if there is a faster method, to which another participant suggests trying the substitution u=x-2 on the second integral. The first participant then asks for guidance on how to proceed with this substitution, receiving advice to carry on with their initial attempt.
  • #1
Clara Chung
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Homework Statement


integrate 1/(x+x^3+2)

Homework Equations

The Attempt at a Solution


I have tried to use partial fraction but the process is very complicated. Are there any faster methods? It is one of the ten questions in a 50 minutes elementary level test.

Here is my attempt,
∫ 1/(x^3+x+2) dx= ∫ 1/[4(x+1)] dx - ∫ (x-2)/[4(x^2-x+2)] dx
= In(x+1)/4 - ∫ (x-2)/[4((x-1/2)^2+7/4)] dx + C
Then I have to use complicated trigonometry by letting x - 1/2 = (root 7) tan(y) /2

Please tell me if there is a faster method
 
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  • #2
Don't think there's a faster method. Push forward with what you have!
 
  • #3
Try the substitution ##u=x-2## on the second integral.
 
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  • #4
vela said:
Try the substitution ##u=x-2## on the second integral.
Please teach me how to do after letting u = x-2
∫ (x-2)/[(x^2-x+2)] dx
= ∫ (u / (u^2 + 3u + 4)) du
=?
 
  • #5
Clara Chung said:
Please teach me how to do after letting u = x-2
∫ (x-2)/[(x^2-x+2)] dx
= ∫ (u / (u^2 + 3u + 4)) du
=?

What do YOU think you should do next? What have you tried so far? (I mean besides the ##x\; \text{to}\;y## transformation you already mentioned and that you said you would like to avoid.)
 
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  • #6
Clara Chung said:
Please teach me how to do after letting u = x-2
∫ (x-2)/[(x^2-x+2)] dx
= ∫ (u / (u^2 + 3u + 4)) du
=?
I made a mistake when thinking of how to do the integral, so I don't think this substitution really helps. As Ackbach advised earlier, just carry on with your initial attempt.
 
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  • #7
So so you are doing it by stages and the second stage getting something like

$$\int \dfrac {\left( x-2\right) dx}{\left[ 4\left( x-\dfrac {1}{2}\right) ^{2}+\dfrac {7}{4}\right] } $$

(I have not checked and am not sure of the placing of the 4 and brackets from your text, but the following holds in any case).

So that is a linear (first degree) function divided by a quadratic (second degree). For the next (I am afraid not the last) step - in calculus what is the relation between some first and a second degree polynomial? :oldwink:
 
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1. What is the purpose of integrating 1/(x+x^3+2)?

The purpose of integrating 1/(x+x^3+2) is to find the area under the curve of the given function. Integration is a mathematical concept that allows us to calculate the total area between a curve and the x-axis.

2. What are the steps involved in integrating 1/(x+x^3+2)?

The steps involved in integrating 1/(x+x^3+2) are as follows:

  1. Perform partial fraction decomposition to simplify the function.
  2. Use the power rule to integrate each term of the decomposed function.
  3. Add a constant of integration to the final solution.

3. Is there a specific method to integrate 1/(x+x^3+2)?

Yes, there are various methods to integrate 1/(x+x^3+2) such as using partial fraction decomposition, substitution, or integration by parts. The choice of method depends on the complexity of the function and the desired level of accuracy.

4. Can the integral of 1/(x+x^3+2) be evaluated using a calculator?

Yes, most scientific calculators have a function to evaluate integrals. However, it is important to note that the result may not always be accurate and it is recommended to double check the solution using other methods.

5. What are the applications of integrating 1/(x+x^3+2)?

Integrating 1/(x+x^3+2) has various applications in physics, engineering, and economics. For example, it can be used to calculate the work done by a varying force, the displacement of a moving object, or the total cost of production in a business. It is a fundamental tool in solving real-world problems that involve finding the area under a curve.

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