Question

\[\int\sqrt{3 x^2 + 4x + 1}\text{  dx }\]

Answer 1

\[\int\sqrt{3 x^2 + 4x + 1} \text{  dx }\]
\[ = \sqrt{3}\int\sqrt{x^2 + \frac{4}{3}x + \frac{1}{3}}\text{  dx }\]
\[ = \sqrt{3}\int\sqrt{x^2 + \frac{4}{3}x + \left( \frac{2}{3} \right)^2 - \left( \frac{2}{3} \right)^2 + \frac{1}{3}} \text{  dx }\]
\[ = \sqrt{3}\int\sqrt{\left( x + \frac{2}{3} \right)^2 - \frac{4}{9} + \frac{1}{3}} \text{  dx }\]
\[ = \sqrt{3}\int\sqrt{\left( x + \frac{2}{3} \right)^2 - \left( \frac{1}{3} \right)^2} \text{  dx }\]
\[ = \sqrt{3} \left[ \frac{1}{2}\left( x + \frac{2}{3} \right)\sqrt{\left( x + \frac{2}{3} \right)^2 - \left( \frac{1}{3} \right)^2} - \frac{1}{2} \times \left( \frac{1}{3} \right)^2 \text{ ln } \left| \left( x + \frac{2}{3} \right) + \sqrt{\left( x + \frac{2}{3} \right)^2 - \left( \frac{1}{3} \right)^2} \right| + C \right] ....................\left[ \because \int \sqrt{x^2 - a^2} dx = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{1}{2} a^2 \text{ ln 
}\left| x + \sqrt{x^2 - a^2} \right| + C \right]\]
\[ = \frac{1}{6}\left( 3x + 2 \right)\sqrt{3 x^2 + 4x + 1} - \frac{\sqrt{3}}{18}\text{ ln } \left| \left( x + \frac{2}{3} \right) + \sqrt{x^2 + \frac{4}{3}x + \frac{1}{3}} \right| + C\]