Question

\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]

 

Answer 1

\[\text{ We  have,} \]
\[I = \int \left( \frac{x^2 + 9}{x^4 + 81} \right)dx\]
\[\text{Dividing numerator and denominator by} \text{ x}^2 \]
\[I = \int\frac{\left( 1 + \frac{9}{x^2} \right)dx}{x^2 + \frac{81}{x^2}}\]
\[ = \int\frac{\left( 1 + \frac{9}{x^2} \right)dx}{x^2 + \left( \frac{9}{x} \right)^2 - 2 \times x \times \frac{9}{x} + 2 \times x \times \frac{9}{x}}\]
\[ = \int\frac{\left( 1 + \frac{9}{x^2} \right)dx}{\left( x - \frac{9}{x} \right)^2 + \left( \sqrt{18} \right)^2}\]
\[\text{ Putting x }- \frac{9}{x} = t\]
\[ \Rightarrow \left( 1 + \frac{9}{x^2} \right)dx = dt\]
\[ \therefore I = \int\frac{dt}{t^2 + \left( \sqrt{18} \right)^2}\]
\[ = \int\frac{dt}{t^2 + \left( 3\sqrt{2} \right)^2}\]
\[ = \frac{1}{3\sqrt{2}} \tan^{- 1} \left( \frac{t}{3\sqrt{2}} \right) + C\]
\[ = \frac{1}{3\sqrt{2}} \tan^{- 1} \left( \frac{x - \frac{9}{x}}{3\sqrt{2}} \right) + C\]
\[ = \frac{1}{3\sqrt{2}} \tan^{- 1} \left( \frac{x^2 - 9}{3\sqrt{2}x} \right) + C\]