Question

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

Answer 1

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