Question

\[\int\frac{1}{x^4 + 3 x^2 + 1} \text{ dx }\]

Answer 1

\[\text{ We have}, \]
\[I = \int \frac{dx}{x^4 + 3 x^2 + 1}\]
\[ = \frac{1}{2}\int \frac{2 \text{ dx }}{x^4 + 3 x^2 + 1}\]
\[ = \frac{1}{2}\int\left[ \frac{\left( x^2 + 1 \right) - \left( x^2 - 1 \right)}{x^4 + 3 x^2 + 1} \right]dx\]
\[ = \frac{1}{2}\int\left( \frac{x^2 + 1}{x^4 + 3 x^2 + 1} \right)dx - \frac{1}{2}\int\frac{\left( x^2 - 1 \right)}{x^4 + 3 x^2 + 1}dx\]
\[\text{Dividing numerator and denominator by} \text{ x}^2 \]
\[ = \frac{1}{2}\int\left( \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2} + 3} \right)dx - \frac{1}{2}\int\frac{\left( 1 - \frac{1}{x^2} \right)dx}{x^2 + \frac{1}{x^2} + 3}\]
\[ = \frac{1}{2}\int\left( \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2} - 2 + 5} \right)dx - \frac{1}{2}\int\frac{\left( 1 - \frac{1}{x^2} \right)dx}{x^2 + \frac{1}{x^2} + 2 + 1}\]
\[ = \frac{1}{2}\int\frac{\left( 1 + \frac{1}{x^2} \right)dx}{\left( x - \frac{1}{x} \right)^2 + \left( \sqrt{5} \right)^2} - \frac{1}{2}\int\frac{\left( 1 - \frac{1}{x^2} \right)dx}{\left( x + \frac{1}{x} \right)^2 + 1^2}\]
\[\text{ Putting  x} - \frac{1}{x} = t\]
\[ \Rightarrow \left( 1 + \frac{1}{x^2} \right)dx = dt\]
\[\text{ Putting  x} + \frac{1}{x} = p\]
\[ \Rightarrow \left( 1 - \frac{1}{x^2} \right)dx = dp\]
\[ \therefore I = \frac{1}{2}\int\frac{dt}{t^2 + \left( \sqrt{5} \right)^2} - \frac{1}{2}\int\frac{dp}{p^2 + 1^2}\]
\[ = \frac{1}{2\sqrt{5}} \tan^{- 1} \left( \frac{t}{\sqrt{5}} \right) - \frac{1}{2} \tan^{- 1} \left( p \right) + C\]
\[ = \frac{1}{2\sqrt{5}} \tan^{- 1} \left( \frac{x - \frac{1}{x}}{\sqrt{5}} \right) - \frac{1}{2} \tan^{- 1} \left( x + \frac{1}{x} \right) + C\]
\[ = \frac{1}{2\sqrt{5}} \tan^{- 1} \left( \frac{x^2 - 1}{\sqrt{5}x} \right) - \frac{1}{2} \tan^{- 1} \left( \frac{x^2 + 1}{x} \right) + C\]