text: # How to evaluate $\int \frac{x^6}{x^4-1} \, \mathrm{d}x.$
  
  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!
  
  -
  First do division, then factor the denominator using difference of squares. – Adam Hughes Jul 1 '14 at 20:48
  
  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.
  
  -
  When I did the log division I got 1 + (x^2/x^4 - 1) o_o where did I go wrong? – user152099 Jul 1 '14 at 20:53
  You long division is correct. – amWhy Jul 1 '14 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 '14 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 '14 at 21:07
  You made a mistake. – Tunk-Fey Jul 2 '14 at 10:02
  
  Integrand can be transformed without long divison $$\begin{gathered}\frac{x^6}{x^4 - 1}=\frac{x^6-1+1}{x^4 - 1}=\frac{(x^2-1)(x^4+x^2+1)}{x^4 - 1}+\frac{1}{x^4 - 1}=\\ =\frac{x^4+x^2+1}{x^2 + 1} + \frac{1}{x^4 - 1}=\\ =x^2+\frac{1}{x^2 + 1}+\frac{1}{2}\left(\frac{1}{x^2-1} - \frac{1}{x^2+1} \right)=\\ =x^2 + \frac{1}{2}\left(\frac{1}{x^2-1} + \frac{1}{x^2+1} \right). \end{gathered}$$
  
  -
  
  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
  
  -
explanation: The document contains a math question and a detailed discussion on how to solve it. The question is about evaluating the integral \(\int \frac{x^6}{x^4-1} \, \mathrm{d}x\). The user has provided their progress and is seeking help to solve for the constants \(A\), \(B\), \(C\), and \(D\) in the partial fraction decomposition. The document also includes comments and a detailed solution from another user.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # How to evaluate $\int \frac{x^6}{x^4-1} \, \mathrm{d}x.$
  
  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!
  
  -
  First do division, then factor the denominator using difference of squares. – Adam Hughes Jul 1 '14 at 20:48
  
  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.
  
  -
  When I did the log division I got 1 + (x^2/x^4 - 1) o_o where did I go wrong? – user152099 Jul 1 '14 at 20:53
  You long division is correct. – amWhy Jul 1 '14 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 '14 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 '14 at 21:07
  You made a mistake. – Tunk-Fey Jul 2 '14 at 10:02
  
  Integrand can be transformed without long divison $$\begin{gathered}\frac{x^6}{x^4 - 1}=\frac{x^6-1+1}{x^4 - 1}=\frac{(x^2-1)(x^4+x^2+1)}{x^4 - 1}+\frac{1}{x^4 - 1}=\\ =\frac{x^4+x^2+1}{x^2 + 1} + \frac{1}{x^4 - 1}=\\ =x^2+\frac{1}{x^2 + 1}+\frac{1}{2}\left(\frac{1}{x^2-1} - \frac{1}{x^2+1} \right)=\\ =x^2 + \frac{1}{2}\left(\frac{1}{x^2-1} + \frac{1}{x^2+1} \right). \end{gathered}$$
  
  -
  
  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
  
  -
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed discussion on how to solve it. The question is about evaluating the integral \(\int \frac{x^6}{x^4-1} \, \mathrm{d}x\). The user has provided their progress and is seeking help to solve for the constants \(A\), \(B\), \(C\), and \(D\) in the partial fraction decomposition. The document also includes comments and a detailed solution from another user.
  
  Label: question-and-answer