Question

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

Answer 1

\[\text{ Let I } = \int \left( x + 1 \right) \sqrt{2 x^2 + 3} \text{ dx}\]
\[\text{ Also, x} + 1 = \lambda\frac{d}{dx}\left( 2 x^2 + 3 \right) + \mu\]
\[ \Rightarrow x + 1 = \lambda\left( 4x \right) + \mu\]
\[\text{Equating coefficient of like terms}\]
\[4\lambda = 1\]
\[ \Rightarrow \lambda = \frac{1}{4} \text{ and }\mu = 1\]
\[ \therefore I = \int \left[ \frac{1}{4}\left( 4x \right) + 1 \right] \sqrt{2 x^2 + 3} \text{ dx}\]
\[ = \frac{1}{4}\int \left( 4x \right) \sqrt{2 x^2 + 3} \text{ dx}+ \int\sqrt{2 x^2 + 3} \text{ dx}\]
\[ = \frac{1}{4}\int\left( 4x \right)\sqrt{2 x^2 + 3}\text{ dx}+ \int\sqrt{2\left( x^2 + \frac{3}{2} \right)}\text{ dx}\]
\[ = \frac{1}{4}\int\left( 4x \right) \sqrt{2 x^2 + 3} \text{ dx}+ \sqrt{2} \int\sqrt{x^2 + \left( \frac{\sqrt{3}}{\sqrt{2}} \right)^2} \text{ dx}\]
\[\text{ Let 2 x}^2 + 3 = t\]
\[ \Rightarrow 4x \text{ dx}= dt\]
\[ \therefore I = \frac{1}{4}\int \sqrt{t}\text{  dt} + \sqrt{2}\left[ \frac{x}{2}\sqrt{x^2 + \frac{3}{2}} + \frac{3}{4}\text{ log }\left| x + \sqrt{x^2 + \frac{3}{2}} \right| \right]\]
\[ = \frac{1}{4}\left[ \frac{t^\frac{3}{2}}{\frac{3}{2}} \right] + \sqrt{2}\frac{x}{2}\frac{\sqrt{2 x^2 + 3}}{\sqrt{2}} + \frac{3\sqrt{2}}{4}\text{ log } \left| x + \frac{\sqrt{2 x^2 + 3}}{\sqrt{2}} \right| + C\]
\[ = \frac{1}{6} \left( 2 x^2 + 3 \right)^\frac{3}{2} + \frac{x}{2}\sqrt{2 x^2 + 3} + \frac{3\sqrt{2}}{4}\text{ log } \left| \frac{\sqrt{2}x + \sqrt{2 x^2 + 3}}{\sqrt{2}} \right| + C\]
\[ = \frac{1}{6} \left( 2 x^2 + 3 \right)^\frac{3}{2} + \frac{x}{2}\sqrt{2 x^2 + 3} + \frac{3\sqrt{2}}{4}\text{ log }\left| \sqrt{2}x + \sqrt{2 x^2 + 3} \right| - \frac{3\sqrt{2}}{4}\text{ log }\sqrt{2} + C\]
\[ = \frac{1}{6} \left( 2 x^2 + 3 \right)^\frac{3}{2} + \frac{x}{2}\sqrt{2 x^2 + 3} + \frac{3\sqrt{2}}{4}\text{ log }\left| \sqrt{2}x + \sqrt{2 x^2 + 3} \right| + C'\]
\[\text{ Where C' = C} - \frac{3\sqrt{2}}{4}\text{ log } \sqrt{2}\]