Question

\[\int\sqrt{x^2 + x + 1} \text{ dx}\]

Answer 1

\[\int \sqrt{x^2 + x + 1} \text{ dx}\]
\[ = \int \sqrt{x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 1} \text{ dx}\]
\[ = \int \sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[ = \left( \frac{x + \frac{1}{2}}{2} \right) \sqrt{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} + \frac{3}{8}\text{ log } \left| \left( x + \frac{1}{2} \right) + \frac{1}{2} + \sqrt{x^2 + x + 1} \right| + C \left[ \because \int\sqrt{x^2 + a^2}dx = \frac{1}{2}x\sqrt{x^2 + a^2} + \frac{1}{2} a^2 \text{ ln }\left| x + \sqrt{x^2 + a^2} \right| + C \right]\]
\[ = \left( \frac{2x + 1}{4} \right) \sqrt{x^2 + x + 1} + \frac{3}{8}\text{ log }\left| \left( 2x + 1 \right) + \frac{1}{2} + \sqrt{x^2 + x + 1} \right| + C\]