Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\displaystyle \int\frac{1}{1+x^8}dx$

My Try:: First we will factorise $1+x^8 = 1^2+(x^4)^2+2x^4-2x^4$

$ = (1+x^4)^2-(\sqrt{2}.x^2)^2 = (x^4+\sqrt{2}x^2+1).(x^4-\sqrt{2}x^2+1)$

So $\displaystyle \int\frac{1}{1+x^8}dx = \int \frac{1}{(x^4+\sqrt{2}x^2+1).(x^4-\sqrt{2}x^2+1)}dx$

Now for partial fraction solving Let we will take $x^2 = t$

$\displaystyle \frac {1}{(t^2+\sqrt{2}t+1).(t^2-\sqrt{2}t+1)} = \frac{At+B}{t^2+\sqrt{2}t+1}+\frac{Ct+D}{t^2-\sqrt{2}t+1}$

But Using This Method solution steps are very Complex

So anyone have a Nice Method to solve Given Question.

Thanks

share|improve this question
1  
By the looks of this, I don't think we'll be able to find a nice method to solve this integral. –  Jared May 22 '13 at 4:43
    
@Jared I was curious when there was a step by step solution button xD –  DanZimm May 22 '13 at 4:45
    
What is interesting to me is that there appears to be a nice structure in the integral of $\frac{1}{1+x^n}$ for any $n$ you like. Try, for example, integrals.wolfram.com/… for $n=111$. –  Lord Soth May 22 '13 at 4:47
    
As an aside, $$\int_0^\infty\frac{dx}{1+x^n}=\frac{\frac\pi n}{\sin\left(\frac\pi n\right)}$$ –  Lucian Dec 9 '13 at 2:08

1 Answer 1

Why not splitting up in fractions until you have first degree polynomials in the nominators?

$$\frac{1}{1+x^8}=\frac{A}{x-e^{i\pi/8}}+\frac{B}{x-e^{-i\pi/8}}+\frac{C}{x-e^{i3\pi/8}}+\frac{D}{x-e^{-i3\pi/8}}+\frac{E}{x-e^{i5\pi/8}}+\frac{F}{x-e^{-i5\pi/8}}+\frac{G}{x-e^{i7\pi/8}}+\frac{H}{x-e^{-i7\pi/8}}$$

or if you prefer without the complex numbers

$$\frac{1}{1+x^8}=\frac{ax+b}{x^2-2\cos(\pi/8)x+1}+\frac{cx+d}{x^2-2\cos(3\pi/8)x+1}+\frac{ex+f}{x^2-2\cos(5\pi/8)x+1}+\frac{gx+h}{x^2-2\cos(7\pi/8)x+1} \; .$$

With the complex formula, you can find the coefficients easily as follows

$$A=\lim_{x \to e^{i\pi/8}}\frac{x-e^{i\pi/8}}{1+x^8}\overset{\text{H}}{=}\lim_{x \to e^{i\pi/8}}\frac{1}{8x^7}=\frac{e^{-i7\pi/8}}{8}$$

where I used de l'Hôpital's rule.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.