Question

Integrate the following function: f(x)=$\frac{1}{x^4+x^2+1}$

Answer 1

$x^4+x^2+1=(x^4+2x^2+1)-x^2=(x^2+1{)}^2-x^2$

$\Rightarrow x^4+x^2+1=(x^2+x+1)(x^2-x+1)$

$\Rightarrow \int \frac{1}{x^4+x^2+1}dx=\int \frac{1}{(x^2+x+1)(x^2-x+1)}dx$

$\Rightarrow \frac{1}{2}\int \frac{(x^3+1)-(x^3-1)}{(x^2+x+1)(x^2-x+1)}dx$

$\Rightarrow \frac{1}{2}\int \frac{x^3+1}{(x^2+x+1)(x^2-x+1)}dx-\frac{1}{2}\int \frac{x^3-1}{(x^2+x+1)(x^2-x+1)}dx$

$\Rightarrow \frac{1}{2}\int \frac{x+1}{x^2+x+1}dx-\frac{1}{2}\int \frac{x-1}{x^2-x+1}dx$

$\Rightarrow \frac{1}{4}\int \frac{2x+2}{x^2+x+1}dx-\frac{1}{4}\int \frac{2x-2}{x^2-x+1}dx$

$\Rightarrow \frac{1}{4}\int \frac{2x+1+1}{x^2+x+1}dx-\frac{1}{4}\int \frac{2x-1-1}{x^2-x+1}dx$

$\Rightarrow \frac{1}{4}\int \frac{2x+1}{x^2+x+1}dx+\frac{1}{4}\int \frac{1}{(x+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}dx-\frac{1}{4}\int \frac{2x-1}{x^2-x+1}dx+\frac{1}{4}\int \frac{1}{(x-\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}dx$

$\Rightarrow \frac{1}{4}\int \frac{d(x^2+x+1)}{x^2+x+1}+\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)-\frac{1}{4}\int \frac{d(x^2-x+1)}{x^2-x+1}+\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$

$\Rightarrow \frac{1}{4}\log |x^2+x+1|-\frac{1}{4}\log |x^2-x+1|+\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$

We know that $ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)$

$\Rightarrow \frac{1}{4}\log |x^2+x+1|-\frac{1}{4}\log |x^2-x+1|+\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{\sqrt{3}x}{1-x^2}\right)+C$

$\Rightarrow \frac{1}{4}\log |\frac{x^2+x+1}{x^2-x+1}|+\frac{1}{2\sqrt{3}}ta{n}^{-1}\left(\frac{\sqrt{3}x}{1-x^2}\right)+C$