Question

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

Answer 1

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