Find: (3x + 1) 4 - 3x - 2x^ 2 dx - Vidyadip

Question

Find: $\int(3x + 1)\sqrt{4 - 3x - 2x^{2 } }\text{ dx}$

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Answer

$\int(3\text{x} + 1)\sqrt{4 - 3\text{x} - 2\text{x}^{2}}\text{ dx} = -\frac{3}{4}\int(-4\text{x} - 3)\sqrt{4 - \text{3x} - \text{2x}^{2}} \text{ dx} - \frac{5}{4}\int\sqrt{4 - 3\text{x} - 2\text{x}^{2}} \text{ dx}$
$= -\frac{1}{2}(4 - \text{3x} - \text{2x}^{2})^{3/2}-\frac{5}{4}\sqrt{2}\int\sqrt{\bigg(\frac{\sqrt{41}}{4}\bigg)^{2} - \bigg(\text{x} + \frac{3}{4}\bigg)^{2}} \text{dx}$
$= -\frac{1}{2}(4 - \text{3x} - \text{2x}^{2})^{3/2}-\frac{5}{4}\sqrt{2} \left\{\frac{4\text{x} + 3}{8} \sqrt{\frac{41}{16}- \bigg(\text{x} + \frac{3}{4}\bigg)^{2}} + \frac{41}{32}.\sin^{-1}\bigg(\frac{\text{4x + 3}}{\sqrt{41}}\bigg)\right\} + \text{C}$
$= -\frac{1}{2}(4 - \text{3x} - \text{2x}^{2})^{3/2}-\frac{5}{4}\left\{\frac{\text{4x + 3}}{8} \sqrt{4 - \text{3x - 2x}^{2}} + \frac{41\sqrt{2}}{32}.\sin^{-1}\bigg(\frac{\text{4x + 3}}{\sqrt{41}}\bigg)\right\} + \text{C}$

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