Question

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

Answer 1

` ∫ {x  dx}/{x^4 - x^2 + 1}`
\[\text{ Let } x^2 = t\]
\[ \Rightarrow \text{ 2x dx } = dt\]
\[ \Rightarrow \text{ x dx } = \frac{dt}{2}\]
Now, ` ∫ {x  dx}/{x^4 - x^2 + 1}`
\[ = \frac{1}{2}\int\frac{dt}{t^2 - t + 1}\]
\[ = \frac{1}{2}\int\frac{dt}{t^2 - t + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 1}\]
\[ = \frac{1}{2}\int\frac{dt}{\left( t - \frac{1}{2} \right)^2 + \frac{3}{4}}\]
\[ = \frac{1}{2}\int\frac{dt}{\left( t - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[ = \frac{1}{2} \times \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{t - \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{2t - 1}{\sqrt{3}} \right) + C\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{2 x^2 - 1}{\sqrt{3}} \right) + C\]