Question

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

Answer 1

We have,

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

Putting `x^2 = t`

\[\text{Then, }\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} = \frac{t}{\left( t + 1 \right) \left( 3t + 4 \right)}\]

\[\text{Let }\frac{t}{\left( t + 1 \right) \left( 3t + 4 \right)} = \frac{A}{t + 1} + \frac{B}{3t + 4}\]

\[ \Rightarrow \frac{t}{\left( t + 1 \right) \left( 3t + 4 \right)} = \frac{A \left( 3t + 4 \right) + B \left( t + 1 \right)}{\left( t + 1 \right) \left( 3t + 4 \right)}\]

\[ \Rightarrow t = A \left( 3t + 4 \right) + B \left( t + 1 \right)\]

Putting `t + 1 = 0`

\[ \Rightarrow t = - 1\]

\[ \therefore - 1 = A \left( - 3 + 4 \right) + 0\]

\[ \Rightarrow A = - 1\]

Putting `3t + 4 = 0`

\[ \Rightarrow t = - \frac{4}{3}\]

\[ \therefore - \frac{4}{3} = 0 + B \left( - \frac{4}{3} + 1 \right)\]

\[ \Rightarrow - \frac{4}{3} = B \times \left( - \frac{1}{3} \right)\]

\[ \Rightarrow B = 4\]

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

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

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

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

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

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