Question

\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]

Answer 1

\[ \text{ Let I} = \int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]
\[\text{ let 2x + 3 = A}\frac{d}{dx}\left( 4 x^2 + 5x + 6 \right) + B\]
\[ \Rightarrow 2x + 3 = A \left( 8x + 5 \right) + B . . . (1)\]
\[\text{By equating coefficients of like terms we get}, \]
\[\text{ 2x = 8A x }\]
\[ \Rightarrow A = \frac{1}{4}\]
\[ \text{ and  5A + B = 3}\]
\[ \Rightarrow \frac{5}{4} + B = 3\]
\[ \Rightarrow B = 3 - \frac{5}{4}\]
\[ = \frac{7}{4}\]
\[\text{Thus, by substituting the values of A and B in eq (1) we ge}t\]
\[I = \int \left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]
\[ = \int\left[ \frac{1}{4}\left( 8x + 5 \right) + \frac{7}{4} \right] \sqrt{4 x^2 + 5x + 6} \text{ dx}\]
\[ = \frac{1}{4}\int\left( 8x + 5 \right) \sqrt{4 x^2 + 5x + 6} dx + \frac{7}{4} \int\sqrt{4 x^2 + 5x + 6} \text{ dx}\]
\[Putting\ 4 x^2 + 5x + 6 = \text{    t   in the first integral}\]
\[ \Rightarrow \left( 8x + 5 \right) \text{ dx}= dt\]
\[ \therefore I = \frac{1}{4}\int\sqrt{t} \cdot dt + \frac{7 \times 2}{4}\int\sqrt{x^2 + \frac{5x}{4} + \frac{3}{2}} \text{ dx}\]
\[ = \frac{1}{4}\int t^\frac{1}{2} \cdot dt + \frac{7}{2}\int\sqrt{x^2 - \frac{5x}{4} + \left( \frac{5}{8} \right)^2 - \left( \frac{5}{8} \right)^2 + \frac{3}{2}} \text{ dx}\]
\[ = \frac{1}{4} \left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + \frac{7}{2}\int\sqrt{\left( x + \frac{5}{8} \right)^2 - \frac{25}{64} + \frac{3}{2}} \text{ dx}\]
\[ = \frac{1}{4} \times \frac{2}{3} t^\frac{3}{2} + \frac{7}{2}\int\sqrt{\left( x + \frac{5}{8} \right)^2 + \frac{- 25 + 96}{64}}\]
\[ = \frac{1}{6} t^\frac{3}{2} + \frac{7}{2}\int\sqrt{\left( x + \frac{5}{8} \right)^2 + \left( \frac{\sqrt{71}}{8} \right)^2}\]
\[ = \frac{1}{6} \left( 4 x^2 + 5x + 6 \right)^\frac{3}{2} + \frac{7}{2}\left[ \frac{x + \frac{5}{8}}{2}\sqrt{\left( x + \frac{5}{8} \right)^2 + \left( \frac{\sqrt{71}}{8} \right)^2} + \frac{71}{64 \times 2} \text{ ln} \left| x + \frac{5}{8} + \sqrt{\left( x + \frac{5}{8} \right)^2 + \left( \frac{\sqrt{71}}{8} \right)^2} \right| \right] + C ................\left[ \because \int\sqrt{a^2 + x^2} \text{ dx}= \frac{1}{2}x\sqrt{a^2 + x^2} + \frac{1}{2} a^2 \text{ ln}\left| x + \sqrt{x^2 + a^2} \right| + C \right]\]
\[ = \frac{1}{6} \left( 4 x^2 + 5x + 6 \right)^\frac{3}{2} + \frac{7}{2} \frac{\left( 8x + 5 \right)}{16} \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} + \frac{71 \times 7}{2 \times 128} \text{ ln} \left| x + \frac{5}{8} + \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} \right| + C\]
\[ = \frac{1}{6} \left( 4 x^2 + 5x + 6 \right)^\frac{3}{2} + \frac{7 \times 2 \left( 8x + 5 \right)}{4 \times 16} \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} + \frac{497}{256} \text{ ln} \left| x + \frac{5}{8} + \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} \right| + C\]
\[ = \frac{1}{6} \left( 4 x^2 + 5x + 6 \right) \sqrt{4 x^2 + 5x + 6} + \frac{7}{64} \left( 8x + 5 \right) \sqrt{4 x^2 + 5x + 6} + \frac{497}{256} \text{ ln }\left| x + \frac{5}{6} + \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} \right| + C\]
\[ = \sqrt{4 x^2 + 5x + 6} \left[ \frac{4 x^2 + 5x + 6}{6} + \frac{7}{64} \left( 8x + 5 \right) \right] + \frac{497}{256} \text{ ln} \left| x + \frac{5}{8} + \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} \right| + C\]
\[ = \sqrt{4 x^2 + 5x + 6} \left[ \frac{128 x^2 + 328x + 297}{192} \right] + \text{ ln} \left| x + \frac{5}{8} + \sqrt{x^2 + \frac{5}{4}x + \frac{3}{2}} \right| + C\]