Integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1)

In summary, you can integrate the function y = \int_{0}^{1} 1/(u^5+1)du using the method outlined in the previous section.
  • #1
DH
21
0
Determine the integral:
[tex]y = \int_{0}^{1} 1/(u^4+1)du [/tex]
and
[tex]y = \int_{0}^{1} 1/(u^5+1)du [/tex]
and
[tex]y = \int_{0}^{1} 1/(u^6+1)du [/tex]
 
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  • #2
Just factor the denominator (to quadratics) and use partial fractions.
 
  • #3
Data said:
Just factor the denominator (to quadratics) and use partial fractions.

no no, can't do like that!
 
  • #4
don't see why not~
 
  • #5
Data said:
don't see why not~
show me the solution!
 
  • #6
[tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

[tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]

[tex]= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)[/tex]

and from there it's just minor substitutions to finish.
 
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  • #7
Hix, you did the wrong thing form row 1 -> row 2!
 
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  • #8
Data said:
[tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

[tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]

[tex]= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)[/tex]

and from there it's just minor substitutions to finish.
how about this:
[tex]y = \int_{0}^{1} 1/(u^5+1)du [/tex]
 
  • #9
Same advice. You can do it yourself this time :wink:
 
  • #10
Data said:
Same advice. You can do it yourself this time :wink:
but one more time, can you show me the solution! I really want to know the way you solve it to compare with my own method... Please give me the solution in detail!
 
  • #11
You can do it almost exactly the same way I did the last one - factor the denominator (in this case, it factors to a product of two irreducible quadractics and the linear factor (x+1)), then use partial fractions to separate it into a sum of functions that you know how to integrate.

Why don't you post an example of a solution using your method? I'm interested now!
 
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  • #12
Data said:
[tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

[tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]

[tex]= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)[/tex]

and from there it's just minor substitutions to finish.
You made a mistake here:
[tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

[tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]
 
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  • #13
indeed I did, the second line should be

[tex]\int_0^1 \frac{du}{2}\left( \frac{\frac{1}{\sqrt{2}}u + 1}{u^2+\sqrt{2}u+1} - \frac{\frac{1}{\sqrt{2}}u - 1}{u^2-\sqrt{2}u+1}\right),[/tex]

but the simplification after that is still similar, just with more terms.
 
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Related to Integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1)

1. What is the purpose of integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1)?

The purpose of integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1) is to find the area under the curve of these functions. Integration is a mathematical tool used to find the total value of a function over a given interval. By integrating these functions, we can determine their total value and use this information in various applications such as physics, engineering, and economics.

2. What is the general method for integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1)?

The general method for integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1) is to use substitution. This involves replacing the variable u with a new variable, usually denoted by t, and then applying the appropriate integration techniques. In this case, we would use the substitution u = tan(t) to solve for the integral.

3. Are there any special cases or exceptions when integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1)?

Yes, there are special cases or exceptions when integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1). For example, if the degree of the denominator is higher than the degree of the numerator, we may need to use partial fractions to solve the integral. Additionally, if the function has singularities or discontinuities within the interval of integration, this can also affect the integration process.

4. What are some real-world applications of integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1)?

Integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1) has various real-world applications. For example, it can be used to calculate the volume of certain shapes, such as a torus or a sphere. It is also used in physics to determine the work done by a variable force and in economics to calculate the total revenue or profit of a business.

5. How does integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1) relate to other mathematical concepts?

Integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1) is closely related to other mathematical concepts such as derivatives, limits, and series. Integration is the inverse process of differentiation, and it involves finding the antiderivative of a function. It also relies on the concept of limits to determine the area under a curve. Additionally, it is used in the study of infinite series, where the integral of a function is equivalent to the sum of an infinite number of terms.

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