Question

\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]

Answer 1

\[I = \int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\]
\[\text{ Let } \left( 2x + 5 \right) = A\frac{d}{dx}\left( 10 - 4x - 3 x^2 \right) + B\]
\[ \Rightarrow \left( 2x + 5 \right) = A\left( - 4 - 6x \right) + B\]
\[ \Rightarrow \left( 2x + 5 \right) = - 6Ax + \left( B - 4A \right)\]
\[ \Rightarrow 2 = - 6A\text{  and } \left( B - 4A \right) = 5\]
\[ \Rightarrow A = - \frac{1}{3} \text{ and B }= \frac{11}{3}\]

\[\Rightarrow \left( 2x + 5 \right) = - \frac{1}{3}\left( - 4 - 6x \right) + \frac{11}{3}\]
\[ \Rightarrow I = - \frac{1}{3}\int\left( - 4 - 6x \right)\sqrt{10 - 4x - 3 x^2}dx + \frac{11}{3}\int\sqrt{10 - 4x - 3 x^2}dx\]
\[\text{ Let I }= - \frac{1}{3} I_1 + \frac{11}{3} I_2 . . . \left( i \right)\]
\[\text{ Now,} \]
\[ I_1 = \int\left( - 4 - 6x \right)\sqrt{10 - 4x - 3 x^2}dx\]
\[\text{ Let }\left( 10 - 4x - 3 x^2 \right) = t, or, \left( - 4 - 6x \right)dx = dt\]
\[ \Rightarrow I_1 = \int\sqrt{t}dt\]
\[ = \frac{2}{3} t^\frac{3}{2} + c_1 \]
\[ \Rightarrow I_1 = \frac{2}{3} \left( 10 - 4x - 3 x^2 \right)^\frac{3}{2} + c_1\]

\[I_2 = \int\sqrt{\left( 10 - 4x - 3 x^2 \right)}dx\]
\[ = \int\sqrt{3\left( \frac{10}{3} - \frac{4}{3}x - x^2 \right)}dx\]
\[ = \sqrt{3}\int\sqrt{\left( \frac{26}{9} - \frac{4}{9} - \frac{4}{3}x - x^2 \right)}dx\]
\[ = \sqrt{3}\int\sqrt{\left[ \left( \frac{\sqrt{26}}{3} \right)^2 - \left( \frac{4}{9} + \frac{4}{3}x + x^2 \right) \right]}dx\]
\[ = \sqrt{3}\int\sqrt{\left[ \left( \frac{\sqrt{26}}{3} \right)^2 - \left( x + \frac{2}{3} \right)^2 \right]}dx\]
\[ = \sqrt{3}\sin\left( \frac{x + \frac{2}{3}}{\frac{\sqrt{26}}{3}} \right) + c_2 \]
\[ = \sqrt{3}\sin\left( \frac{3x + 2}{\sqrt{26}} \right) + c_2\]

Using (i), we get

\[I = - \frac{1}{3} \times \frac{2}{3} \left( 10 - 4x - 3 x^2 \right)^\frac{3}{2} + \frac{11}{3} \times \sqrt{3}\sin\left( \frac{3x + 2}{\sqrt{26}} \right) + C\]
\[ \therefore I = - \frac{2}{9} \left( 10 - 4x - 3 x^2 \right)^\frac{3}{2} + \frac{11\sqrt{3}}{3}\sin\left( \frac{3x + 2}{\sqrt{26}} \right) + C\]