Question

\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]

Answer 1

We have,

\[I = \int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]

Putting `x^2 = t`

\[\text{Then, }\frac{1}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)} = \frac{1}{\left( t + 1 \right) \left( t + 4 \right)}\]

\[\text{Let }\frac{1}{\left( t + 1 \right) \left( t + 4 \right)} = \frac{A}{t + 1} + \frac{B}{t + 4}\]

\[ \Rightarrow 1 = A \left( t + 4 \right) + B \left( t + 1 \right)\]

Putting `t + 4 = 0`

\[ \Rightarrow t = - 4\]

\[ \therefore 1 = A \times 0 + B \left( - 3 \right)\]

\[ \Rightarrow B = - \frac{1}{3}\]

Putting `t + 1 = 0`

\[ \Rightarrow t = - 1\]

\[ \therefore 1 = A \left( - 1 + 4 \right) + B \times 0\]

\[ \Rightarrow A = \frac{1}{3}\]

\[ \therefore \frac{1}{\left( t + 1 \right) \left( t + 4 \right)} = \frac{1}{3 \left( t + 1 \right)} - \frac{1}{3 \left( t + 4 \right)}\]

\[ \Rightarrow \frac{1}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)} = \frac{1}{3 \left( x^2 + 1 \right)} - \frac{1}{3 \left( x^2 + 2^2 \right)}\]

\[ \Rightarrow \int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)} = \frac{1}{3}\int\frac{dx}{x^2 + 1^2} - \frac{1}{3}\int\frac{dx}{x^2 + 2^2}\]

\[ = \frac{1}{3} \tan^{- 1} x - \frac{1}{3} \times \frac{1}{2} \tan^{- 1} \left( \frac{x}{2} \right) + C\]

\[ = \frac{1}{3} \tan^{- 1} x - \frac{1}{6} \tan^{- 1} \left( \frac{x}{2} \right) + C\]