Question

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

Answer 1

$$\begin{gathered} \int \frac{xdx}{x^4-x^2+1} \\ \mathrm{Let}{x}^2=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \mathrm{Now},\int \frac{xdx}{x^4-x^2+1} \\ =\frac{1}{2}\int \frac{dt}{t^2-t+1} \\ =\frac{1}{2}\int \frac{dt}{t^2-t+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2+1} \\ =\frac{1}{2}\int \frac{dt}{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2+{\displaystyle \frac{3}{4}}} \\ =\frac{1}{2}\int \frac{dt}{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ =\frac{1}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{t-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)+C \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)+C \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x^2-1}{\sqrt{3}}\right)+C \end{gathered}$$