Question

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

Answer 1

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