cool integral

DH

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]

Data

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

DH

no no, can't do like that!

Data

don't see why not~

DH

show me the solution!

Data

[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.

DH

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

DH

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

Data

Same advice. You can do it yourself this time

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!

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!

DH

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]

Data

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.