anonymous
  • anonymous
integrate x/(81x^4-1)
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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misty1212
  • misty1212
HI!
misty1212
  • misty1212
factor the denominator, then go with partial fractions
misty1212
  • misty1212
\[81x^4-1=(3x+1)(3x-1)(9x^2+1)\] is the start

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misty1212
  • misty1212
you know how to do partial fraction decomposition?
anonymous
  • anonymous
Yes! I did it, I just can't seem to get the correct answer
mathmale
  • mathmale
Could you possibly share your work, so that we could give you meaningful feedback on what you have done? Thanks.
mathmale
  • mathmale
I agree with Misty's expansion of 81x^4-1. The partial fraction expansion should have the form:
mathmale
  • mathmale
\[\frac{ x }{ (3x+1)(3x-1)(9x^2+1) }=\frac{ A }{ 3x+1 }+\frac{ B }{ 3x-1 }+\frac{ Cx+D }{ 9x^2+1 }\]
mathmale
  • mathmale
Your job is to determine the values of A, B, C and D.
anonymous
  • anonymous
my answer was \[\ln \left| \frac{ 9x^2-1 }{ 9x^2+1 } \right| + C\]
SolomonZelman
  • SolomonZelman
Integrals like this can be also solved this way, ((for example)) \(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{x}{25x^4-4}~{\rm d}x}\) \(\large\color{#8080ff}{\displaystyle u=x^2}\) \(\large\color{#8080ff}{\displaystyle {\rm d}u=2x~{\rm d}x\quad \Longrightarrow \quad {\rm d}u/2=x~{\rm d}x}\) \(\large\color{#000000}{\displaystyle\frac{1}{2}\int\limits_{~}^{~}\frac{1}{25u^2-4}~{\rm d}u}\) \(\large\color{#000000}{\displaystyle\frac{1}{50}\int\limits_{~}^{~}\frac{1}{u^2-\frac{4}{25}}~{\rm d}u}\) \(\large\color{#000000}{\displaystyle\frac{1}{50}\int\limits_{~}^{~}\frac{1}{u^2-\left[\frac{2}{5}\right]^2}~{\rm d}u}\) \(\large\color{#8080ff}{ u= \frac{2}{5}\csc\theta}\) \(\large\color{#8080ff}{ {\rm d}u=-\frac{2}{5}\cos\theta\csc^2\theta~{\rm d}\theta }\) \(\large\color{#000000}{\displaystyle\frac{1}{50}\int\limits_{~}^{~}\frac{-\frac{2}{5}\cos\theta\csc^2\theta}{\left[\frac{2}{5}\csc\theta\right]^2-\left[\frac{2}{5}\right]^2}~{\rm d}\theta }\) Note that cscĀ²(w)-1=cotĀ²(w) \(\large\color{#000000}{\displaystyle\frac{1}{50}\int\limits_{~}^{~}\frac{-\frac{2}{5}\cos\theta\csc^2\theta}{\left[\frac{2}{5}\right]^2\cot^2\theta}~{\rm d}\theta }\) \(\large\color{#000000}{\displaystyle\frac{1}{50}\int\limits_{~}^{~}\frac{-\cos\theta\csc^2\theta}{\frac{2}{5}\cot^2\theta}~{\rm d}\theta }\) \(\large\color{#000000}{\displaystyle\frac{-1}{20}\int\limits_{~}^{~}\frac{\cos\theta\csc^2\theta}{\cot^2\theta}~{\rm d}\theta }\) \(\large\color{#000000}{\displaystyle\frac{-1}{20}\int\limits_{~}^{~}\frac{\cos\theta\csc^2\theta}{\cos^2\theta \csc^2\theta }~{\rm d}\theta }\) \(\large\color{#000000}{\displaystyle\frac{-1}{20}\int\limits_{~}^{~}\sec \theta }\)
SolomonZelman
  • SolomonZelman
and then of course the formula for secant (which can also be derived through u-sub if you want)
mathmale
  • mathmale
Thank you, Solomon, for suggesting this alternative method ()trig substitution).

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