Question

integrate x/(81x^4-1)

Answer 1

HI!

Answer 2

factor the denominator, then go with partial fractions

Answer 3

\[81x^4-1=(3x+1)(3x-1)(9x^2+1)\] is the start

Answer 4

you know how to do partial fraction decomposition?

Answer 5

Yes! I did it, I just can't seem to get the correct answer

Answer 6

Could you possibly share your work, so that we could give you meaningful feedback on what you have done? Thanks.

Answer 7

I agree with Misty's expansion of 81x^4-1.

The partial fraction expansion should have the form:

Answer 8

\[\frac{ x }{ (3x+1)(3x-1)(9x^2+1) }=\frac{ A }{ 3x+1 }+\frac{ B }{ 3x-1 }+\frac{ Cx+D }{ 9x^2+1 }\]

Answer 9

Your job is to determine the values of A, B, C and D.

Answer 10

my answer was \[\ln \left| \frac{ 9x^2-1 }{ 9x^2+1 } \right| + C\]

Answer 11

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 csc²(w)-1=cot²(w)

\(\large\color{#000000}{\displaystyle\frac{1}{50}\int\limits_{~}^{~}\frac{-\frac{2}{5}\cos\theta\csc^2\theta}{\left[\frac{2}{5}\right]^2\cot^2\theta}~{\rm d}\theta }\)

\(\large\color{#000000}{\displaystyle\frac{1}{50}\int\limits_{~}^{~}\frac{-\cos\theta\csc^2\theta}{\frac{2}{5}\cot^2\theta}~{\rm d}\theta }\)

\(\large\color{#000000}{\displaystyle\frac{-1}{20}\int\limits_{~}^{~}\frac{\cos\theta\csc^2\theta}{\cot^2\theta}~{\rm d}\theta }\)

\(\large\color{#000000}{\displaystyle\frac{-1}{20}\int\limits_{~}^{~}\frac{\cos\theta\csc^2\theta}{\cos^2\theta \csc^2\theta }~{\rm d}\theta }\)

\(\large\color{#000000}{\displaystyle\frac{-1}{20}\int\limits_{~}^{~}\sec \theta }\)

Answer 12

and then of course the formula for secant (which can also be derived through u-sub if you want)

Answer 13

Thank you, Solomon, for suggesting this alternative method ()trig substitution).