Question

Find : \[\int\frac{\left( x^2 + 1 \right)\left( x^2 + 4 \right)}{\left( x^2 + 3 \right)\left( x^2 - 5 \right)}dx\] .

Answer 1

\[\int\frac{\left( x^2 + 1 \right)\left( x^2 + 4 \right)}{\left( x^2 + 3 \right)\left( x^2 - 5 \right)}dx\]

\[\text { Let } x^2 = t\]

\[ \therefore \frac{\left( x^2 + 1 \right)\left( x^2 + 4 \right)}{\left( x^2 + 3 \right)\left( x^2 - 5 \right)} = \frac{\left( t + 1 \right)\left( t + 4 \right)}{\left( t + 3 \right)\left( t - 5 \right)} = \frac{t^2 + 5t + 4}{\left( t + 3 \right)\left( t - 5 \right)} = 1 + \frac{7t + 19}{\left( t + 3 \right)\left( t - 5 \right)}\]

\[\text { Let } \frac{7t + 19}{\left( t + 3 \right)\left( t - 5 \right)} = \frac{A}{t + 3} + \frac{B}{t - 5}\]

\[ \Rightarrow 7t + 19 = A\left( t - 5 \right) + B\left( t + 3 \right)\]

\[\text { Putting }t = 5, \text { we get } B = \frac{27}{4}\]

\[\text { Putting } t = - 3, \text { we get } A = \frac{1}{4}\]

\[ \therefore \frac{t^2 + 5t + 4}{\left( t + 3 \right)\left( t - 5 \right)} = 1 + \frac{1}{4\left( t + 3 \right)} + \frac{27}{4\left( t - 5 \right)}\]

\[ \Rightarrow \int\frac{\left( x^2 + 1 \right)\left( x^2 + 4 \right)}{\left( x^2 + 3 \right)\left( x^2 - 5 \right)}dx = \int dx + \frac{1}{4}\int\frac{1}{\left( x^2 + 3 \right)}dx + \frac{27}{4}\int\frac{1}{\left( x^2 - 5 \right)}dx\]

\[ = x + \frac{1}{4 \times \sqrt{3}} \tan^{- 1} \left( \frac{x}{\sqrt{3}} \right) + \frac{27}{4} \times \frac{1}{2\sqrt{5}}\log\left| \frac{x - \sqrt{5}}{x + \sqrt{5}} \right| + C\]

\[ = x + \frac{1}{4\sqrt{3}} \tan^{- 1} \left( \frac{x}{\sqrt{3}} \right) + \frac{27}{8\sqrt{5}}\log\left| \frac{x - \sqrt{5}}{x + \sqrt{5}} \right| + C\]