# Cool integral

by DH
Tags: integral
 P: 22 Determine the integral: $$y = \int_{0}^{1} 1/(u^4+1)du$$ and $$y = \int_{0}^{1} 1/(u^5+1)du$$ and $$y = \int_{0}^{1} 1/(u^6+1)du$$
 P: 998 Just factor the denominator (to quadratics) and use partial fractions.
P: 22
 Quote by Data Just factor the denominator (to quadratics) and use partial fractions.

no no, can't do like that!

 P: 998 Cool integral don't see why not~
P: 22
 Quote by Data don't see why not~
show me the solution!
 P: 998 $$\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}$$ $$= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)$$ $$= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)$$ and from there it's just minor substitutions to finish.
 P: 22 Hix, you did the wrong thing form row 1 -> row 2!!!!
P: 22
 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.
$$y = \int_{0}^{1} 1/(u^5+1)du$$
 P: 998 Same advice. You can do it yourself this time
P: 22
 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!
 P: 998 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!
P: 22
 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.
$$\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}$$
$$= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)$$
 P: 998 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.