<s>[INST] Integral of quartic function in denominator I'm sorry, I've really tried to use MathJaX but I can't get integrals to work properly.
indefinite integral
$$\int {x\over x^4 +x^2 +1}$$
I set it up to equal
$$x\int {x\over x^4 +x^2 +1} - \int {x\over x^4 +x^2 +1}$$
$$\text{so } (x-1)\int {1\over x^4 +x^2 +1}$$
OKAY, now I set the denominator to $(X^2 +.5)^2 + \frac 34$
So I multiplied the top and bottom by $\frac 43$
then I absorbed it into the squared quantity by dividing it (within the parenthesis) by $\sqrt 3\over 2$
so
$${1\over \left({X^2 +.5\over {\sqrt 3\over 2}}\right)^2 + 1}$$
so $\arctan\left({X^2 +.5\over{\sqrt 3\over 2}}\right)$
FINAL ANSWER: $(x-1)\arctan\left({X^2 +.5\over{\sqrt 3\over 2}}\right) + C$
Thanks for reading, I have no way of checking this work... tutors are always have too many people wanting help.
 [/INST] HINT:
As $\displaystyle x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$ and $\displaystyle (x^2+x+1)-(x^2-x+1)=2x$
$$\frac x{x^4+x^2+1}=\frac12\frac{(x^2+x+1)-(x^2-x+1)}{x^4+x+1}=\frac12\left(\frac1{x^2-x+1}-\frac1{x^2+x+1}\right)$$
Now  as, $\displaystyle x^2+x+1=\frac{(2x+1)^2+(\sqrt3)^2}4$ put $2x+1=\sqrt3\tan\theta$
and similarly for $\displaystyle\int\frac{dx}{x^2-x+1}$ 
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