Question

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

Answer 1

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