cool integral

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

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

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

no no, can't do like that!

cool integral

don't see why not~

 Quote by Data don't see why not~
show me the solution!

 $$\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}$$ $$= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)$$ $$= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)$$ and from there it's just minor substitutions to finish.
 Hix, you did the wrong thing form row 1 -> row 2!!!!

 Quote by 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.
$$y = \int_{0}^{1} 1/(u^5+1)du$$

 Same advice. You can do it yourself this time

 Quote by Data Same advice. You can do it yourself this time
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!

 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!

 Quote by 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.
$$\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}$$
$$= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)$$
 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.