Question

\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]

Answer 1

\[\text{ We have,} \]
\[I = \int \left( \frac{x^2 + 1}{x^4 + 7 x^2 + 1} \right)dx\]
\[\text{Dividing numerator and denominator by} \text{ x}^2 \]
\[I = \int\left( \frac{1 + \frac{1}{x^2}}{x^2 + 7 + \frac{1}{x^2}} \right)dx\]
\[ = \int\frac{\left( 1 + \frac{1}{x^2} \right)dx}{x^2 + \frac{1}{x^2} - 2 + 9}\]
\[ \Rightarrow \int\frac{\left( 1 + \frac{1}{x^2} \right)dx}{\left( x - \frac{1}{x} \right)^2 + 3^2}\]
\[\text{ Putting x} - \frac{1}{x} = t\]
\[ \Rightarrow \left( 1 + \frac{1}{x^2} \right)dx = dt\]
\[ \therefore I = \int\frac{dt}{t^2 + 3^2}\]
\[ = \frac{1}{3} \tan^{- 1} \left( \frac{t}{3} \right) + C\]
\[ = \frac{1}{3} \tan^{- 1} \left( \frac{x - \frac{1}{x}}{3} \right) + C\]
\[ = \frac{1}{3} \tan^{- 1} \left( \frac{x^2 - 1}{3x} \right) + C\]