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]
[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.
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!
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]
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.