Primitive of $\frac{3x^4-1}{(x^4+x+1)^2}$

Question

How to find primitive of: $$\frac{3x^4-1}{(x^4+x+1)^2}$$ I am having a faint idea of a type which may or maynot be in the primitve, i.e.: $$\frac{p(x)}{x^4+x+1}$$ The problem is I am not getting an idea of a substitution to solve this problem.

I might show my work but it is totally useless, atleast in this case.

For reference: enter image description here

Apply partial fraction decomposition, then integrate each term. — Jean-Claude Arbaut, Aug 22 '14 at 19:44
@Jean-ClaudeArbaut to this too exists an easy solution by substitution[that I don't know], which becomes lengthier to calculate via partial fraction, — RE60K, Aug 22 '14 at 19:47
You won't escape using logarithm and arctangent in the answer, though. — Jean-Claude Arbaut, Aug 22 '14 at 19:49
@Tunk-Fey As often, WA does it the wrong way, using complex numbers where they are really useless: Maxima gives an answer involving only real logarithms and arctangents. However, it can't be simpler than that. Rather easy with partial fractions, though cumbersome. — Jean-Claude Arbaut, Aug 22 '14 at 19:52
computers are not as intelligent as human, they can't do everything like humans. they can't innovate new substitutions except the one taught to them, that's why this question is suited to us — RE60K, Aug 22 '14 at 19:55
btw http://www.wolframalpha.com/input/?i=diffrentiate+-x%2F%28x%5E4%2Bx%2B1%29 — RE60K, Aug 22 '14 at 19:55
@Aditya You can innovate as much as you want, there will be logarithms and arctangent in the answer, period. — Jean-Claude Arbaut, Aug 22 '14 at 19:56
@Jean-ClaudeArbaut Ahh, how stupid am I!? Thanks for the link :) — Tunk-Fey, Aug 22 '14 at 19:56
@Jean-ClaudeArbaut Is there a guideline for such serious typos? — RE60K, Aug 22 '14 at 20:02
With the change in the question, everything is easy, but any little change, like changing the $1$ on top to $2$, turns it into a nightmare. — André Nicolas, Aug 22 '14 at 20:05

score 1 · Answer 1 · answered Aug 23 '14 at 14:20

$$ \int \frac{3x^4-1}{(x^4+x+1)^2} = \int \frac{3x^4+4x^3-4x^3-1}{(x^4+x+1)^2}$$ $$ \int \frac{3x^4+4x^3-4x^3-1}{(x^4+x+1)^2} = \int \frac{3x^4+4x^3}{(x^4+x+1)^2}- \int \frac{4x^3+1}{(x^4+x+1)^2}$$ Consider, $$ \int \frac{3x^4+4x^3}{(x^4+x+1)^2} = \int \frac{4x^3(x+1)-x^4}{(x^4+x+1)^2} = \int \frac{(4x^3+1)(x+1)+(-1)(x^4+x+1)}{(x^4+x+1)^2}= -\frac{x+1}{x^4+x+1}$$ Hint: Can you see quotient rule here?

And , obviously $$ \int \frac{4x^3+1}{(x^4+x+1)^2} = -\frac{1}{x^4+x+1} $$ Hint: Use substitution $t=x^4+x+1$

So concluding, $$ \int \frac{3x^4-1}{(x^4+x+1)^2} = -\frac{x}{x^4+x+1} \Box$$

score 1 · Accepted Answer · answered Aug 24 '14 at 05:09

1

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \int\frac{3x^4-1}{x^2\cdot \left(x^3+1+x^{-1}\right)^2}dx = \int\frac{(3x^2-x^{-2})}{(x^3+1+x^{-1})^2}dx$$

Now Let $$(x^3+1+x^{-1}) = t\;,$$ Then $$(3x^2-x^{-2})dx = dt$$

So $$\displaystyle I = \int\frac{1}{t^2}dt = -\frac{1}{t}+\mathbb{C} = -\frac{1}{x^3+1+x^{-1}}+\mathbb{C} = -\left(\frac{x}{x^4+x+1}\right)+\mathbb{C}$$

answered Aug 24 '14 at 05:09

juantheron

53,015

what I was looking for. – RE60K Aug 24 '14 at 05:23

Primitive of $\frac{3x^4-1}{(x^4+x+1)^2}$

2 Answers2