Question

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

Answer 1

\[\text{ We  have,} \]
\[I = \int \frac{dx}{x^4 + x^2 + 1}\]
\[ = \frac{1}{2}\int \frac{2 \text{ dx }}{x^4 + x^2 + 1}\]
\[ \Rightarrow \frac{1}{2}\int\left( \frac{\left( x^2 + 1 \right) - \left( x^2 - 1 \right)}{x^4 + x^2 + 1} \right)dx\]
\[ \Rightarrow \frac{1}{2}\int\left( \frac{x^2 + 1}{x^4 + x^2 + 1} \right)dx - \frac{1}{2}\int\left( \frac{x^2 - 1}{x^4 + x^2 + 1} \right)dx\]
\[\text{Dividing numerator and denominator by} \text{ x}^2 \]
\[I = \frac{1}{2}\int\left( \frac{1 + \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} \right)dx - \frac{1}{2}\int\left( \frac{1 - \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2} - 2 + 3} \right)dx - \frac{1}{2}\int\frac{\left( 1 - \frac{1}{x^2} \right)dx}{x^2 + \frac{1}{x^2} + 2 - 1}\]
\[ = \frac{1}{2}\int\frac{\left( 1 + \frac{1}{x^2} \right)dx}{\left( x - \frac{1}{x} \right)^2 + \left( \sqrt{3} \right)^2} - \frac{1}{2}\int\frac{\left( 1 - \frac{1}{x^2} \right)dx}{\left( x + \frac{1}{x} \right)^2 - 1^2}\]
\[\text{ Putting x } - \frac{1}{x} = t\]
\[ \Rightarrow \left( 1 + \frac{1}{x^2} \right)dx = dt\]
\[\text{ Putting x} + \frac{1}{x} = p\]
\[ \Rightarrow \left( 1 - \frac{1}{x^2} \right)dx = dp\]
\[ \therefore I = \frac{1}{2}\int\frac{dt}{t^2 + \left( \sqrt{3} \right)^2} - \frac{1}{2}\int\frac{dp}{p^2 - 1^2}\]
\[ = \frac{1}{2} \times \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{t}{\sqrt{3}} \right) - \frac{1}{2} \times \frac{1}{2 \times 1}\text{ log }\left| \frac{p - 1}{p + 1} \right| + C\]
\[ = \frac{1}{2\sqrt{3}} \tan^{- 1} \left( \frac{x - \frac{1}{x}}{\sqrt{3}} \right) - \frac{1}{4}\text{ log } \left| \frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1} \right| + C\]
\[ = \frac{1}{2\sqrt{3}} \tan^{- 1} \left( \frac{x^2 - 1}{x\sqrt{3}} \right) - \frac{1}{4}\text{ log }\left| \frac{x^2 - x + 1}{x^2 + x + 1} \right| + C\]