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.

Evaluate the integral:

$$\int \frac{x^6}{x^4-1} \, \mathrm{d}x$$

After a lot of help I have reached this point:

$x^2 = Ax^3 - Ax + Bx^2 - B + Cx^3 + Cx^2 + Cx + C + Dx^3 - Dx^2 + Dx - D$

But now I don't really know how to solve for $A, B, C$, and $D$. Please help!

share|improve this question
1  
First do division, then factor the denominator using difference of squares. –  Adam Hughes Jul 1 at 20:48

3 Answers 3

First: Divide! Use polynomial division to get $$\frac{x^6}{x^4 - 1} = 1 + \frac{x^2}{x^4 - 1}$$ $\int 1\,dx = x + C$

For the second term:

Now factor the denominator, and decompose: $$x^4 - 1 = (x^2 + 1)(x^2-1) = (x^2 + 1)(x-1)(x+1)$$

So the set up we want for the second term is:

$$\frac{x^2}{x^4 - 1} = \dfrac{Ax + B}{x^2 + 1} +\frac{C}{x-1} + \frac D{x+1}$$

Now solve for A, B, C, D.

share|improve this answer
    
When I did the log division I got 1 + (x^2/x^4 - 1) o_o where did I go wrong? –  user152099 Jul 1 at 20:53
    
You long division is correct. –  amWhy Jul 1 at 20:54
    
Would I have to simplify it further like this?: x^2 = Ax+B(x-1)(x+1) + C (x^2+1)(x+1) + D (x^2 + 1)(x-1) and then distribute? –  user152099 Jul 1 at 21:00
    
That would give me x^2 = Ax^3 - Ax + Bx^2 - B + Cx^3 + Cx^2 + Cx + C + Dx^3 - Dx^2 + Dx - D. Now what should I do to solve for A, B, C, and D? I don't really know how to do that. –  user152099 Jul 1 at 21:07
1  
You made a mistake. –  Tunk-Fey Jul 2 at 10:02

Integrand can be transformed without long divison $$\begin{gathered}\frac{x^6}{x^4 - 1}=\frac{x^6-1+1}{x^4 - 1}=\frac{(x^2-1)(x^4+x^2+1)}{x^4 - 1}+\frac{1}{x^4 - 1}=\\ =\frac{x^4+x^2+1}{x^2 + 1} + \frac{1}{x^4 - 1}=\\ =x^2+\frac{1}{x^2 + 1}+\frac{1}{2}\left(\frac{1}{x^2-1} - \frac{1}{x^2+1} \right)=\\ =x^2 + \frac{1}{2}\left(\frac{1}{x^2-1} + \frac{1}{x^2+1} \right). \end{gathered}$$

share|improve this answer

Rewrite the integrand as \begin{align} \frac{x^6}{x^4-1}&\stackrel{\color{red}{[1]}}=\color{darkgreen}{x^2}+\color{blue}{\frac{x^2}{x^4-1}}\\ &=\color{darkgreen}{x^2}+\color{blue}{\frac{x^2}{(x^2-1)(x^2+1)}}\\ &\stackrel{\color{red}{[2]}}=\color{darkgreen}{x^2}+\color{blue}{\frac{1}{2}\left(\color{black}{\frac{1}{x^2+1}}+\color{red}{\frac{1}{x^2-1}}\right)}\\ &=\color{darkgreen}{x^2}+\color{blue}{\frac{1}{2}\left(\color{black}{\frac{1}{x^2+1}}+\color{red}{\frac{1}{(x-1)(x+1)}}\right)}\\ &\stackrel{\color{red}{[3]}}=\color{darkgreen}{x^2}+\color{blue}{\frac{1}{2}\left(\color{black}{\frac{1}{x^2+1}}+\color{red}{\frac12\left[\frac1{x-1}-\frac1{x+1}\right]}\right)}\\ &=x^2+\color{red}{\underbrace{\color{black}{\frac{1}{2(x^2+1)}}}_{\color{blue}{\large \color{black}{\text{set}}\ x=\tan\theta}}}+\color{red}{\underbrace{\color{black}{\frac1{4(x-1)}}}_{\color{blue}{\large \color{black}{\text{set}}\ u=x-1}}}-\color{red}{\underbrace{\color{black}{\frac1{4(x+1)}}}_{\color{blue}{\large \color{black}{\text{set}}\ v=x+1}}} \end{align}


Notes :

$\color{red}{[1]}\;\;\;$Polynomial long division

$\color{red}{[2]}\text{ and }\color{red}{[3]}\;\;\;$Partial fractions decomposition

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.