Question

\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]

Answer 1

\[\text{We have}, \]
\[I = \int\frac{dx}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)}\]
\[\text{ Putting x}^2 = t\]
\[ \therefore \frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} = \frac{1}{\left( t + 2 \right) \left( t + 5 \right)}\]
\[\text{ Let }\frac{1}{\left( t + 2 \right) \left( t + 5 \right)} = \frac{A}{t + 2} + \frac{B}{t + 5}\]
\[ \Rightarrow \frac{1}{\left( t + 2 \right) \left( t + 5 \right)} = \frac{A \left( t + 5 \right) + B \left( t + 2 \right)}{\left( t + 2 \right) \left( t + 5 \right)}\]
\[ \Rightarrow 1 = A \left( t + 5 \right) + B \left( t + 2 \right)\]
\[\text{ Putting t = - 5}\]
\[ \therefore 1 = B \left( - 5 + 2 \right)\]
\[ \Rightarrow B = - \frac{1}{3}\]
\[\text{ Putting t = - 2}\]
\[ \therefore 1 = A \left( - 2 + 5 \right) + B \times 0\]
\[ \Rightarrow A = \frac{1}{3}\]
\[ \therefore I = \frac{1}{3}\int\frac{dx}{x^2 + 2} - \frac{1}{3}\int\frac{dx}{x^2 + 5}\]
\[ = \frac{1}{3}\int\frac{dx}{x^2 + \left( \sqrt{2} \right)^2} - \frac{1}{3}\int\frac{dx}{x^2 + \left( \sqrt{5} \right)^2}\]
\[ = \frac{1}{3\sqrt{2}} \text{ tan}^{- 1} \left( \frac{x}{\sqrt{2}} \right) - \frac{1}{3\sqrt{5}} \text{ tan}^{- 1} \left( \frac{x}{\sqrt{5}} \right) + C\]