Question

\[\int\frac{x^2 - 3x + 1}{x^4 + x^2 + 1} \text{ dx }\]

Answer 1

\[\text{ We have,} \]
\[I = \int\left( \frac{x^2 - 3x + 1}{x^4 + x^2 + 1} \right)dx\]
\[ = \int\frac{\left( x^2 + 1 \right)dx}{x^4 + x^2 + 1} - 3\int\frac{x \text{ dx}}{x^4 + x^2 + 1} . . . . . \left( 1 \right)\]
\[ = I_1 - 3 I_2 \text{ where I}_1 = \int\frac{\left( x^2 + 1 \right)dx}{x^4 + x^2 + 1}, I_2 = \int\frac{x dx}{x^4 + x^2 + 1}\]
\[ I_1 = \int\left( \frac{x^2 + 1}{x^4 + x^2 + 1} \right)dx\]
\[\text{Dividing numerator and denominator by} \text{ x}^2 \]
\[ I_1 = \int\frac{\left( 1 + \frac{1}{x^2} \right)dx}{x^2 + \frac{1}{x^2} + 1}\]
\[ = \int\frac{\left( 1 + \frac{1}{x^2} \right)dx}{x^2 + \frac{1}{x^2} - 2 + 3}\]
\[ = \int\frac{\left( 1 + \frac{1}{x^2} \right)dx}{\left( x - \frac{1}{x} \right)^2 + \left( \sqrt{3} \right)^2}\]
\[\text{ Let x} - \frac{1}{x} = t\]
\[ \Rightarrow \left( 1 + \frac{1}{x^2} \right)dx = dt\]
\[ \therefore I_1 = \int\frac{dt}{t^2 + \left( \sqrt{3} \right)^2}\]
\[ I_1 = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{t}{\sqrt{3}} \right) + C_1 \]
\[ I_1 = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{x - \frac{1}{x}}{\sqrt{3}} \right) + C_1 . . . . . \left( 2 \right)\]
\[ I_2 = \int\frac{x \text{ dx }}{x^4 + x^2 + 1}\]
\[\text{ Putting  x}^2 = t\]
\[ \Rightarrow 2x\text{ dx } = dt\]
\[ \Rightarrow x \text { dx }= \frac{dt}{2}\]
\[ \therefore I_2 = \frac{1}{2}\int\frac{dt}{t^2 + t + 1}\]
\[ = \frac{1}{2}\int\frac{dt}{t^2 + t + \frac{1}{4} + \frac{3}{4}}\]
\[ = \frac{1}{2}\int\frac{dt}{\left( t + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[ = \frac{1}{\frac{\sqrt{3}}{2}} \times \frac{1}{2}\left[ \tan^{- 1} \left( \frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) \right] + C_2 \]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{2t + 1}{\sqrt{3}} \right) + C_2 \]
\[ I_2 = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{2 x^2 + 1}{\sqrt{3}} \right) + C_2 . . . \left( 3 \right)\]
\[\text{ From  equating} \left( 1 \right), \left( 2 \right) \text{ and } \left( 3 \right) \text{ we have}\]
\[I = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{x - \frac{1}{x}}{\sqrt{3}} \right) + C_1 - 3 \times \left[ \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{2 x^2 + 1}{\sqrt{3}} \right) + C_2 \right]\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{x^2 - 1}{\sqrt{3}x} \right) - \sqrt{3} \tan^{- 1} \left( \frac{2 x^2 + 1}{\sqrt{3}} \right) + \text{ C where C = C}_1 + 3 C_2\]