Question

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

Answer 1

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

\[\text{ Therefore }, \]
\[\frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1} = x + 2 + \frac{3x - 1}{x^2 - x + 1} . . . . . \left( 1 \right)\]
\[\text{ Let }\]
\[3x - 1 = A\frac{d}{dx} \left( x^2 - x + 1 \right) + B\]
\[3x - 1 = A \left( 2x - 1 \right) + B\]
\[3x - 1 = \left( 2A \right) x + B - A\]
\[ \text{Equating  Coefficients  of  like } terms\]
\[2A = 3\]
\[A = \frac{3}{2}\]
\[B - A = - 1\]
\[B - \frac{3}{2} = - 1\]
\[B = \frac{1}{2}\]
\[\int\left( \frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1} \right) dx = \int\left( x + 2 \right) dx + \int\left( \frac{\frac{3}{2} \left( 2x - 1 \right) + \frac{1}{2}}{x^2 - x + 1} \right) dx\]

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