Question

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

Answer 1

We have,
\[I = \int \frac{\left( 4 x^4 + 3 \right)dx}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)}\]
\[\text{Putting }x^2 = t\]
Then,
\[\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} = \frac{4 t^2 + 3}{\left( t + 2 \right) \left( t + 3 \right) \left( t + 4 \right)}\]
\[\text{Let }\frac{4 t^2 + 3}{\left( t + 2 \right) \left( t + 3 \right) \left( t + 4 \right)} = \frac{A}{t + 2} + \frac{B}{t + 3} + \frac{C}{t + 4}\]
\[ \Rightarrow \frac{4 t^2 + 3}{\left( t + 2 \right) \left( t + 3 \right) \left( t + 4 \right)} = \frac{A\left( t + 3 \right) \left( t + 4 \right) + B\left( t + 2 \right) \left( t + 4 \right) + C\left( t + 2 \right) \left( t + 3 \right)}{\left( t + 2 \right) \left( t + 3 \right) \left( t + 4 \right)}\]
\[ \Rightarrow 4 t^2 + 3 = A\left( t + 3 \right) \left( t + 4 \right) + B\left( t + 2 \right) \left( t + 4 \right) + C\left( t + 2 \right) \left( t + 3 \right)\]
\[\text{Putting t + 3 = 0}\]
\[ \Rightarrow t = - 3\]
\[ \therefore 4 \times \left( - 3 \right)^2 + 3 = B\left( - 3 + 2 \right) \left( - 3 + 4 \right)\]
\[ \Rightarrow 39 = B\left( - 1 \right)\]
\[ \Rightarrow B = - 39\]
\[\text{Putting t + 2 = 0}\]
\[ \Rightarrow t = - 2\]
\[ \therefore 4 \left( - 2 \right)^2 + 3 = A\left( - 2 + 3 \right) \left( - 2 + 4 \right)\]
\[ \Rightarrow 19 = A \times 1 \times 2\]
\[ \Rightarrow A = \frac{19}{2}\]
\[\text{Let t + 4 = 0}\]
\[ \Rightarrow t = - 4\]
\[ \therefore 4 \times \left( - 4 \right)^2 + 3 = C\left( - 4 + 2 \right) \left( - 4 + 3 \right)\]
\[ \Rightarrow 67 = C\left( - 2 \right) \left( - 1 \right)\]
\[ \Rightarrow C = \frac{67}{2}\]
\[ \therefore \frac{4 t^2 + 3}{\left( t + 2 \right) \left( t + 3 \right) \left( t + 4 \right)} = \frac{19}{2\left( t + 2 \right)} - \frac{39}{t + 3} + \frac{67}{2\left( t + 4 \right)}\]
\[ \Rightarrow \frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} = \frac{19}{2\left( x^2 + 2 \right)} - \frac{39}{x^2 + 3} + \frac{67}{2\left( x^2 + 4 \right)}\]
\[ \therefore I = \frac{19}{2}\int\frac{dx}{x^2 + \left( \sqrt{2} \right)^2} - 39\int\frac{dx}{x^2 + \left( \sqrt{3} \right)^2} - \frac{67}{2}\int\frac{dx}{x^2 + 2^2}\]
\[ = \frac{19}{2} \times \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{x}{\sqrt{2}} \right) - \frac{39}{\sqrt{3}} \tan^{- 1} \left( \frac{x}{\sqrt{3}} \right) - \frac{67}{2} \times \frac{1}{2} \tan^{- 1} \left( \frac{x}{2} \right) + C\]
\[ = \frac{19}{2\sqrt{2}} \tan^{- 1} \left( \frac{x}{\sqrt{2}} \right) - \frac{39}{\sqrt{3}} \tan^{- 1} \left( \frac{x}{\sqrt{3}} \right) - \frac{67}{4} \tan^{- 1} \left( \frac{x}{2} \right) + C\]