Question

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

Answer 1

\[\int \sqrt{2 x^2 + 3x + 4}\text{ dx}\]
\[ = \sqrt{2} \int \sqrt{x^2 + \frac{3}{2}x + 2} \text{ dx}\]
\[ = \sqrt{2} \int \sqrt{x^2 + \frac{3}{2}x + \left( \frac{3}{4} \right)^2 - \left( \frac{3}{4} \right)^2 + 2} \text{ dx}\]
\[ = \sqrt{2} \int \sqrt{\left( x + \frac{3}{4} \right)^2 - \frac{9}{16} + 2} \text{ dx}\]
\[ = \sqrt{2} \int \sqrt{\left( x + \frac{3}{4} \right)^2 + \left( \frac{\sqrt{23}}{4} \right)^2}\text{ dx}\]
\[ = \sqrt{2} \left[ \frac{x + \frac{3}{4}}{2} \sqrt{\left( x + \frac{3}{4} \right)^2 + \left( \frac{\sqrt{23}}{4} \right)^2} + \frac{23}{32}\text{ ln} \left| x + \frac{3}{4} + \sqrt{x^2 + \frac{3}{2}x + 2} \right| \right] + C \left[ \because \int\sqrt{x^2 + a^2} dx = \frac{1}{2}x\sqrt{x^2 + a^2} - \frac{1}{2} a^2 \text{ ln }\left| x + \sqrt{x^2 + a^2} \right| + C \right]\]
\[ = \sqrt{2} \left[ \left( \frac{4x + 3}{8} \right) \sqrt{x^2 + \frac{3}{2}x + 2} + \frac{23}{32}\text{ ln } \left| x + \frac{3}{4} + \sqrt{x^2 + \frac{3}{2}x + 2} \right| \right] + C\]
\[ = \left( \frac{4x + 3}{8} \right) \sqrt{2 x^2 + 3x + 4} + \frac{23\sqrt{2}}{32}\text{ ln }\left| x + \frac{3}{4} + \sqrt{x^2 + \frac{3}{2}x + 2} \right| + C\]