# Cool integral

1. Apr 29, 2005

### DH

Determine the integral:
$$y = \int_{0}^{1} 1/(u^4+1)du$$
and
$$y = \int_{0}^{1} 1/(u^5+1)du$$
and
$$y = \int_{0}^{1} 1/(u^6+1)du$$

Last edited: Apr 29, 2005
2. Apr 29, 2005

### Data

Just factor the denominator (to quadratics) and use partial fractions.

3. Apr 29, 2005

### DH

:surprised
no no, can't do like that!

4. Apr 29, 2005

### Data

don't see why not~

5. Apr 29, 2005

### DH

show me the solution!

6. Apr 29, 2005

### Data

$$\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}$$

$$= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)$$

$$= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)$$

and from there it's just minor substitutions to finish.

Last edited: Apr 29, 2005
7. Apr 29, 2005

### DH

Hix, you did the wrong thing form row 1 -> row 2!!!!

Last edited: Apr 29, 2005
8. Apr 29, 2005

### DH

how about this:
$$y = \int_{0}^{1} 1/(u^5+1)du$$

9. Apr 29, 2005

### Data

Same advice. You can do it yourself this time

10. Apr 29, 2005

### DH

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. Apr 29, 2005

### Data

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!

Last edited: Apr 29, 2005
12. Apr 29, 2005

### DH

You made a mistake here:
$$\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}$$

$$= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)$$

Last edited: Apr 29, 2005
13. Apr 29, 2005

### Data

indeed I did, the second line should be

$$\int_0^1 \frac{du}{2}\left( \frac{\frac{1}{\sqrt{2}}u + 1}{u^2+\sqrt{2}u+1} - \frac{\frac{1}{\sqrt{2}}u - 1}{u^2-\sqrt{2}u+1}\right),$$

but the simplification after that is still similar, just with more terms.

Last edited: Apr 29, 2005
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