Question

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

Answer 1

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

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

$=\frac{1}{2}\left[\int \frac{(1-\frac{1}{x^2})dx}{x^2+\frac{1}{x^2}+1}+\int \frac{(1+\frac{1}{x^2})dx}{x^2+\frac{1}{x^2}+1}\right]$

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

Take $x+\frac{1}{x}=t$

$\Rightarrow t^2-1=x^2+\frac{1}{x^2}+1$

$(1-\frac{1}{x^2})dx=dt$

$=\int \frac{dt}{t^2-1}=\frac{1}{2\left(1\right)}ln\left|\frac{t-1}{t+1}\right|+c=\frac{1}{2}ln\left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|+c$

$=\frac{1}{2}ln\left|\frac{x^2-x+1}{x^2+x+1}\right|+c$

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

$(1+\frac{1}{x^2})dx=du$ $\Rightarrow u^2+3=x^2+\frac{1}{x^2}+1$

$=\int \frac{du}{u^2+3}=\int \frac{du}{u^2+(\sqrt{3}{)}^2}=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{u}{\sqrt{3}}\right)+c$

$=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+c$

$=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)+c$

$\int \frac{x^2}{x^4+x^2+1}dx=\frac{1}{2}ln\left|\frac{x^2-x+1}{x^2+x+1}\right|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)+c$.