Integral of 1/u^n knowing integral of 1/u

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Hello,

This is the initial problem:

Find the primitive of 1/(x4+1)

i've done this, the value of the primitive is this ugly looking expression:

http://image.bayimg.com/d9ac5052a83d5808955a7a647c6cb7343f9ace1f.jpg

Now the question asked is to deduce the primitive value of 1/(x4+1)^3 from what I found.

This is why I'm asking if there is any general method to compute the value of ∫1/u^n knowing ∫1/u

Thank you
 
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No, essentially there isn't. \int \frac{1}{(1+x^4)^3} is solved through partial fractions.
 
ezintegral said:
Hello,

This is the initial problem:

Find the primitive of 1/(x4+1)

i've done this, the value of the primitive is this ugly looking expression:

http://image.bayimg.com/d9ac5052a83d5808955a7a647c6cb7343f9ace1f.jpg
Your image isn't showing.
ezintegral said:
Now the question asked is to deduce the primitive value of 1/(x4+1)^3 from what I found.

This is why I'm asking if there is any general method to compute the value of ∫1/u^n knowing ∫1/u

Two comments here:
1. The two formulas are not related.
$$ \int \frac{du}{u^n} = \int u^{-n}du = \frac{u^{-n + 1}}{-n + 1} + C$$
$$\int \frac{du}{u} = ln|u| + C$$

The integrals you show don't fit either of these formulas, because if u = x4 + 1, then du = 4x3dx, which you don't have.

2. You have omitted the differential du from your integrals above. This isn't so crucial in the early stages of learning integration, but it is for more complicated methods such as integration by parts and trig substitution.
 
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Thanks for the answers.

I'm going to try rewriting 1/(x4+1) in partial equations and see what I can get.
 
ezintegral said:
Thanks for the answers.

I'm going to try rewriting 1/(x4+1) in partial equations and see what I can get.

Presumably you mean 1/(x4 + 1). At the very least, use '^' to write this as 1/(x^4 + 1).

As for breaking up 1/(x4 + 1) using partial fractions, I don't see how this will do you any good. The denominator cannot be reduced to lower-degree factors with real coefficients.

It would help if you showed us the exact problem you're working on.
 
Mark44 said:
... The denominator cannot be reduced to lower-degree factors with real coefficients.
...
x4 + 1 can be reduced to lower-degree factors with real coefficients, but not to lower-degree factors with integer coefficients.

x^4+1=(x^2+(\sqrt{2})x+1)(x^2-(\sqrt{2})x+1)
 
Mark44 said:
Presumably you mean 1/(x4 + 1). At the very least, use '^' to write this as 1/(x^4 + 1).

As for breaking up 1/(x4 + 1) using partial fractions, I don't see how this will do you any good. The denominator cannot be reduced to lower-degree factors with real coefficients.

It would help if you showed us the exact problem you're working on.

This problem is presented in 2 questions.

1) Find the primitive of \frac{1}{x^4 + 1}

This is done, no particular problem. You get this:

http://image.bayimg.com/d9ac5052a83d5808955a7a647c6cb7343f9ace1f.jpg


2) Deduce the primitive of \frac{1}{(x^4 + 1)^3} from the primitive of \frac{1}{x^4 + 1}


Now the problem is with 2). I don't see how I can use the first result to find the second one.

Thank you for your time.
 
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SammyS said:
x4 + 1 can be reduced to lower-degree factors with real coefficients, but not to lower-degree factors with integer coefficients.

x^4+1=(x^2+(\sqrt{2})x+1)(x^2-(\sqrt{2})x+1)
Thanks...
 
ezintegral said:
This problem is presented in 2 questions.

1) Find the primitive of \frac{1}{x^4 + 1}

This is done, no particular problem. You get this:

http://image.bayimg.com/d9ac5052a83d5808955a7a647c6cb7343f9ace1f.jpg


2) Deduce the primitive of \frac{1}{(x^4 + 1)^3} from the primitive of \frac{1}{x^4 + 1}


Now the problem is with 2). I don't see how I can use the first result to find the second one.

Thank you for your time.
The image above is attached here.
d9ac5052a83d5808955a7a647c6cb7343f9ace1f.jpg
 
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  • #11
dextercioby said:
I_{n} = \int\frac{dx}{\left(1+x^4\right)^n}, \\ ~, n\in \mathbb{N}

Point 1 asks you to compute I1. I3 follows from twice part integration.

Integration by parts once is enough to determine a recursive relationship of the form I_{n+1}(x)=f(x, n)I_n(x)+g(x, n). I_3 can then be expressed in terms of I_1(x) by using that relationship twice.
 
  • #12
gabbagabbahey said:
Integration by parts once is enough to determine a recursive relationship of the form I_{n+1}(x)=f(x, n)I_n(x)+g(x, n). I_3 can then be expressed in terms of I_1(x) by using that relationship twice.

Thank you, this looks promising and within my reach.
 
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