Question

Integrate $\frac{1}{x^4+x^2+1}$

Answer 1

$\int \frac{dx}{1+x^2+x^4}=\frac{1}{4}\int \frac{4dx}{x^4+x^2+1}$
$=\frac{1}{4}[\int \frac{(2-2x^2)dx}{x^4+x^2+1}+\int \frac{(2+2x^2)dx}{x^4+x^2+1}]$
$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$
$=\frac{1}{4}\left[\{\int \frac{2x+1}{x^2+x+1}dx-\int \frac{(2x-1)dx}{x^2-x+1}\}\right]+\frac{1}{4}\left[\{\int \frac{dx}{x^2+x+1}+\int \frac{dx}{x^2-x+1}\}\right]$
$=\frac{1}{4}\left[ln\right(x^2+x+1)-ln(x^2-x+1\left)\right]+\frac{1}{4}[\int \frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+\int \frac{dx}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}]$
$=\frac{1}{4}\left[ln\right(x^2+x+1)-ln(x^2-x+1\left)\right]+\frac{1}{4}[\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)]$
$=\frac{1}{4}ln(x^2+x+1)-\frac{1}{4}ln(x^2-x+1)+\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$