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Cool integralby DH
Tags: integral 
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#1
Apr2905, 01:44 AM

P: 22

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] 


#2
Apr2905, 02:08 AM

P: 998

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



#3
Apr2905, 02:14 AM

P: 22

no no, can't do like that! 


#4
Apr2905, 02:31 AM

P: 998

Cool integral
don't see why not~



#5
Apr2905, 02:41 AM

P: 22




#6
Apr2905, 03:01 AM

P: 998

[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. 


#7
Apr2905, 03:04 AM

P: 22

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



#8
Apr2905, 03:09 AM

P: 22

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


#9
Apr2905, 03:11 AM

P: 998

Same advice. You can do it yourself this time



#10
Apr2905, 03:15 AM

P: 22




#11
Apr2905, 03:18 AM

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! 


#12
Apr2905, 08:38 AM

P: 22

[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] 


#13
Apr2905, 12:04 PM

P: 998

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. 


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