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Question
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Solution
\[\text{ Let I } = \int \left( x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\text{ Also,} x + 1 = \lambda\frac{d}{dx} \left( x^2 + x + 1 \right) + \mu\]
\[ \Rightarrow x + 1 = \lambda\left( 2x + 1 \right) + \mu\]
\[ \Rightarrow x + 1 = \left( 2\lambda \right)x + \lambda + \mu\]
\[\text{Equating coefficient of like terms}\]
\[2\lambda = 1 \]
\[ \Rightarrow \lambda = \frac{1}{2}\]
\[\text{ And }\]
\[\lambda + \mu = 1\]
\[ \Rightarrow \frac{1}{2} + \mu = 1\]
\[ \therefore \mu = \frac{1}{2}\]
\[ \therefore I = \frac{1}{2}\int \left( 2x + 1 \right) \sqrt{x^2 + x + 1}\text{ dx }+ \frac{1}{2}\int\sqrt{x^2 + x + 1} \text{ dx }\]
\[ = \frac{1}{2}\int\left( 2x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }+ \frac{1}{2}\int\sqrt{x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 1} \text{ dx }\]
\[ = \frac{1}{2}\int\left( 2x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }+ \frac{1}{2}\int\sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} dx\]
\[\text{ Let x}^2 + x + 1 = t\]
\[ \Rightarrow \left( 2x + 1 \right)dx = dt\]
\[\text{ Then },\]
\[I = \frac{1}{2}\int\sqrt{t} \text{ dt} + \frac{1}{2}\left[ \frac{x + \frac{1}{2}}{2}\sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} + \frac{3}{8}\text{ log }\left| \left( x + \frac{1}{2} \right) + \sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right| \right] + C\]
\[ = \frac{1}{2} \times \frac{2}{3} t^\frac{3}{2} + \frac{1}{2}\left[ \left( \frac{2x + 1}{4} \right) \sqrt{x^2 + x + 1} + \frac{3}{8}\text{ log }\left| \left( x + \frac{1}{2} \right) + \sqrt{x^2 + x + 1} \right| \right] + C\]
\[ = \frac{1}{3} \left( x^2 + x + 1 \right)^\frac{3}{2} + \frac{1}{2}\left[ \left( \frac{2x + 1}{4} \right) \sqrt{x^2 + x + 1} + \frac{3}{8}\text{ log }\left| \left( x + \frac{1}{2} \right) + \sqrt{x^2 + x + 1} \right| \right] + C\]
