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.
 A: 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 $\displaystyle\int\frac{dx}{x^2-x+1}$ 
A: $$\int\frac{x}{x^4 +x^2 +1}dx=1/2\int\frac{dt}{t^2+t+1}=1/2\int\frac{dt}{(t+1/2)^2+3/4}$$
$$x^2=t,xdx=dt/2$$
A: $$
\int\frac{x}{x^4+x^2+1}dx=\int\frac{x}{x^4+2x^2+1-x^2}dx=\int\frac{x}{\left(x^2+1\right)-x^2}dx=
$$
$$
=\int\frac{x}{\left(x^2+x+1\right)\left(x^2-x+1\right)}dx=\frac{1}{2}\int\frac{\left(x^2+x+1\right)-\left(x^2-x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}dx=
$$
$$
=\frac{1}{2}\left(\int\frac{dx}{x^2-x+1}-\int\frac{dx}{x^2+x+1}\right)=\frac{1}{2}\left(\int\frac{\left(\frac{2x-1}{2}+\frac{1}{2}\right)dx}{x^2-x+1}-\int\frac{\left(\frac{2x+1}{2}-\frac{1}{2}\right)dx}{x^2+x+1}\right)
$$
$$
=\frac{1}{4}\int\frac{(2x-1)dx}{x^2-x+1}+\frac{1}{4}\int\frac{dx}{(x-\frac{1}{2})^2+\frac{3}{4}}-\frac{1}{4}\int\frac{(2x+1)dx}{x^2+x+1}-\frac{1}{4}\int\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}
$$
$$
=\frac{1}{4}\ln|x^2-x+1|-\frac{1}{4}\ln|x^2+x+1|+\frac{1}{4}\cdot\frac{1}{\frac{\sqrt 3}{2}}\arctan \frac{\frac{2x-1}{2}}{\frac{\sqrt 3}{2}}-\frac{1}{4}\cdot\frac{1}{\frac{\sqrt 3}{2}}\arctan \frac{\frac{2x+1}{2}}{\frac{\sqrt 3}{2}}
$$
$$
=\frac{1}{4}\ln\left\vert\frac{x^2-x+1}{x^2+x+1}\right\vert+\frac{1}{2\sqrt 3}\left(\arctan\frac{2x-1}{\sqrt 3}-\arctan\frac{2x+1}{\sqrt 3}\right)
$$
$$
=\frac{1}{4}\ln\left\vert\frac{x^2-x+1}{x^2+x+1}\right\vert+\frac{1}{2\sqrt 3}\arctan\frac{\frac{2x-1}{\sqrt 3}-\frac{2x+1}{\sqrt 3}}{1+\frac{2x-1}{\sqrt 3}\cdot\frac{2x+1}{\sqrt 3}}
$$
$$
=\frac{1}{4}\ln\left\vert\frac{x^2-x+1}{x^2+x+1}\right\vert+\frac{1}{2\sqrt 3}\arctan\left(\frac{-\sqrt 3}{2x^2+1}\right)+C
$$
