Question

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

Answer 1

\[I = \int\left( 3x + 1 \right)\sqrt{4 - 3x - 2 x^2} \text{  dx }\]
\[\text{ Let }\left( 3x + 1 \right) = A\frac{d}{dx}\left( 4 - 3x - 2 x^2 \right) + B\]
\[ \Rightarrow \left( 3x + 1 \right) = A\left( - 3 - 4x \right) + B\]
\[ \Rightarrow \left( 3x + 1 \right) = - 4\text{ Ax} + \left( B - 3A \right)\]
\[ \Rightarrow 3 = - 4\text{ A and }\left( B - 3A \right) = 1\]
\[ \Rightarrow A = - \frac{3}{4} \text{ and B }= - \frac{5}{4}\]

\[\Rightarrow \left( 3x + 1 \right) = - \frac{3}{4}\left( - 3 - 4x \right) - \frac{5}{4}\]
\[ \Rightarrow I = - \frac{3}{4}\int\left( - 3 - 4x \right)\sqrt{4 - 3x - 2 x^2}dx - \frac{5}{4}\int\sqrt{4 - 3x - 2 x^2}dx\]
\[Let I = - \frac{3}{4} I_1 - \frac{5}{4} I_2 . . . \left( i \right)\]
\[\text{ Now, } \]
\[ I_1 = \int\left( - 3 - 4x \right)\sqrt{4 - 3x - 2 x^2}dx\]
\[\text{ Let } \left( 4 - 3x - 2 x^2 \right) = t, or, \left( - 3 - 4x \right)dx = dt\]
\[ \Rightarrow I_1 = \int\sqrt{t}dt\]
\[ = \frac{2}{3} t^\frac{3}{2} + c_1 \]
\[ \Rightarrow I_1 = \frac{2}{3} \left( 4 - 3x - 2 x^2 \right)^\frac{3}{2} + c_1\]

\[I_2 = \int\sqrt{4 - 3x - 2 x^2}dx\]
\[ = \int\sqrt{2\left( 2 - \frac{3}{2}x - x^2 \right)}dx\]
\[ = \sqrt{2}\int\sqrt{\left( \frac{17}{4} - \frac{9}{4} - \frac{3}{2}x - x^2 \right)}dx\]
\[ = \sqrt{2}\int\sqrt{\left[ \left( \frac{\sqrt{17}}{2} \right)^2 - \left( \frac{9}{4} + \frac{3}{2}x + x^2 \right) \right]}dx\]
\[ = \sqrt{2}\int\sqrt{\left[ \left( \frac{\sqrt{17}}{2} \right)^2 - \left( x + \frac{3}{2} \right)^2 \right]}dx\]
\[ = \sqrt{2}\sin\left( \frac{x + \frac{3}{2}}{\frac{\sqrt{17}}{2}} \right) + c_2 \]
\[ = \sqrt{2}\sin\left( \frac{2x + 3}{\sqrt{17}} \right) + c_2\]

Using (i), we get

\[I = - \frac{3}{4} \times \frac{2}{3} \left( 4 - 3x - 2 x^2 \right)^\frac{3}{2} - \frac{5}{4} \times \sqrt{2}\sin\left( \frac{2x + 3}{\sqrt{17}} \right) + C\]
\[ \therefore I = - \frac{1}{2} \left( 4 - 3x - 2 x^2 \right)^\frac{3}{2} - \frac{5\sqrt{2}}{4}\sin\left( \frac{2x + 3}{\sqrt{17}} \right) + C\]