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

1. Dec 9, 2012

### ezintegral

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 [Broken]

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

Last edited by a moderator: May 6, 2017
2. Dec 9, 2012

### dextercioby

No, essentially there isn't. $\int \frac{1}{(1+x^4)^3}$ is solved through partial fractions.

3. Dec 10, 2012

### Staff: Mentor

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.

Last edited by a moderator: May 6, 2017
4. Dec 10, 2012

### ezintegral

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

5. Dec 10, 2012

### Staff: Mentor

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.

6. Dec 10, 2012

### SammyS

Staff Emeritus
x4 + 1 can be reduced to lower-degree factors with real coefficients, but not to lower-degree factors with integer coefficients.

$x^4+1=(x^2+(\sqrt{2})x+1)(x^2-(\sqrt{2})x+1)$

7. Dec 10, 2012

### ezintegral

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 [Broken]

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.

Last edited by a moderator: May 6, 2017
8. Dec 10, 2012

### Staff: Mentor

Thanks...

9. Dec 10, 2012

### Staff: Mentor

The image above is attached here.

Last edited by a moderator: May 6, 2017
10. Dec 10, 2012

### dextercioby

Last edited: Dec 10, 2012
11. Dec 10, 2012

### gabbagabbahey

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. Dec 10, 2012

### ezintegral

Thank you, this looks promising and within my reach.