Question

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

Answer 1

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

\[\text{ Also,} x + 1 = \lambda\frac{d}{dx} \left( x^2 + x + 1 \right) + \mu\]

\[ \Rightarrow x + 1 = \lambda\left( 2x + 1 \right) + \mu\]

\[ \Rightarrow x + 1 = \left( 2\lambda \right)x + \lambda + \mu\]

\[\text{Equating coefficient of like terms}\]

\[2\lambda = 1 \]

\[ \Rightarrow \lambda = \frac{1}{2}\]

\[\text{ And }\]

\[\lambda + \mu = 1\]

\[ \Rightarrow \frac{1}{2} + \mu = 1\]

\[ \therefore \mu = \frac{1}{2}\]

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

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

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

\[\text{ Let x}^2 + x + 1 = t\]

\[ \Rightarrow \left( 2x + 1 \right)dx = dt\]

\[\text{ Then },\]

\[I = \frac{1}{2}\int\sqrt{t} \text{ dt} + \frac{1}{2}\left[ \frac{x + \frac{1}{2}}{2}\sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} + \frac{3}{8}\text{ log  }\left| \left( x + \frac{1}{2} \right) + \sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right| \right] + C\]
\[ = \frac{1}{2} \times \frac{2}{3} t^\frac{3}{2} + \frac{1}{2}\left[ \left( \frac{2x + 1}{4} \right) \sqrt{x^2 + x + 1} + \frac{3}{8}\text{ log }\left| \left( x + \frac{1}{2} \right) + \sqrt{x^2 + x + 1} \right| \right] + C\]
\[ = \frac{1}{3} \left( x^2 + x + 1 \right)^\frac{3}{2} + \frac{1}{2}\left[ \left( \frac{2x + 1}{4} \right) \sqrt{x^2 + x + 1} + \frac{3}{8}\text{ log }\left| \left( x + \frac{1}{2} \right) + \sqrt{x^2 + x + 1} \right| \right] + C\]