Question

\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

Options

• \[ \frac{1}{5x} \left( 4 + \frac{1}{x^2} \right)^{- 5} + C\]

• \[ \frac{1}{5} \left( 4 + \frac{1}{x^2} \right)^{- 5} + C\]

• \[ \frac{1}{10x} \left( \frac{1}{x^2} + 4 \right)^{- 5} + C\]

• \[ \frac{1}{10} \left( \frac{1}{x^2} + 4 \right)^{- 5} + C\]

   

Answer 1

\[ \frac{1}{10} \left( \frac{1}{x^2} + 4 \right)^{- 5} + C\]

 

\[\text{Let }I = \int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]
\[ = \int\frac{x^9}{x^{12} \left( 4 + \frac{1}{x^2} \right)^6}dx\]
\[ = \int\frac{\frac{1}{x^3}}{\left( 4 + \frac{1}{x^2} \right)^6}dx\]
\[\text{Let }\left( 4 + \frac{1}{x^2} \right) = t\]
\[ \text{On differentiating both sides, we get}\]
\[ - \frac{2}{x^3}dx = dt\]
\[ \therefore I = - \frac{1}{2}\int\frac{1}{\left( t \right)^6}dt\]
\[ = - \frac{1}{2}\left( - \frac{1}{5} \right) t^{- 5} + C\]
\[ = \frac{1}{10} \left( 4 + \frac{1}{x^2} \right)^{- 5} + C\]
\[\text{Therefore, }\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx = \frac{1}{10} \left( 4 + \frac{1}{x^2} \right)^{- 5} + C\]