Question

\[\int\left( x - 3 \right)\sqrt{x^2 + 3x - 18} \text{  dx }\]

Answer 1

\[\text{ Let I } = \int\left( x - 3 \right)\sqrt{x^2 + 3x - 18} \text{  dx }\]
\[\text{ We  express x - 3} = A\left( \frac{d}{d x}\left( x^2 + 3x - 18 \right) \right) + B\]
\[x - 3 = A(2x + 3) + B\]
\[\text{Equating the coefficients of x and constants, we get}\]
\[1 = 2A \text{ and }- 3 = 3A + B\]
\[or A = \frac{1}{2} \text{ and B }= - \frac{9}{2} \]
\[ \therefore I = \int\left( \frac{1}{2}\left( 2x + 3 \right) - \frac{9}{2} \right)\sqrt{x^2 + 3x - 18} \text{  dx }\]
\[ = \frac{1}{2}\int\left( 2x + 3 \right)\sqrt{x^2 + 3x - 18} \text{  dx }- \frac{9}{2}\int\sqrt{x^2 + 3x - 18} \text{  dx }\]
\[ = \frac{1}{2} I_1 - \frac{9}{2} I_2 . . . (1)\]
\[\text{ Now,} I_1 = \int\left( 2x + 3 \right)\sqrt{x^2 + 3x - 18} dx\]
\[ \text{ Let x}^2 + 3x - 18 = u\]
\[ \text{On differentiating both sides, we get}\]
\[ \left( 2x + 3 \right)dx = du\]
\[ \therefore I_1 = \int\sqrt{u}du\]
\[ = \frac{2}{3} u^\frac{3}{2} + c_1 \]
\[ = \frac{2}{3} \left( x^2 + 3x - 18 \right)^\frac{3}{2} + c_1 . . . (2)\]
\[\text{ And,} I_2 = \int\sqrt{x^2 + 3x - 18} \text{  dx }\]
\[ = \int\sqrt{x^2 + 3x + \frac{9}{4} - \frac{9}{4} - 18} \text{  dx }\]
\[ = \int\sqrt{\left( x + \frac{3}{2} \right)^2 - \left( \frac{9}{2} \right)^2} dx\]
\[ \text{ Let} \left( x + \frac{3}{2} \right) = u\]
\[ \text{On differentiating both sides, we get}\]
\[ dx = du\]
\[ \therefore I_2 = \int\sqrt{\left( u \right)^2 - \left( \frac{9}{2} \right)^2} du\]
\[ = \frac{u}{2}\sqrt{\left( u \right)^2 - \left( \frac{9}{2} \right)^2} - \frac{1}{2} \left( \frac{9}{2} \right)^2 \text { log }\left| u + \sqrt{\left( u \right)^2 - \left( \frac{9}{2} \right)^2} \right| + c_2 \]
\[ = \frac{x + \frac{3}{2}}{2}\sqrt{\left( x + \frac{3}{2} \right)^2 - \left( \frac{9}{2} \right)^2} - \frac{1}{2} \left( \frac{9}{2} \right)^2 \text{ log}\left| \left( x + \frac{3}{2} \right) + \sqrt{\left( x + \frac{3}{2} \right)^2 - \left( \frac{9}{2} \right)^2} \right| + c_2 \]
\[ = \frac{2x + 3}{4}\sqrt{x^2 + 3x - 18} - \frac{81}{8}\text{ log }\left| \left( x + \frac{3}{2} \right) + \sqrt{x^2 + 3x - 18} \right| + c_2 . . . (3)\]
\[\text{ From (1), (2) and (3), we get }\]
\[ \therefore I = \frac{1}{2}\left( \frac{2}{3} \left( x^2 + 3x - 18 \right)^\frac{3}{2} + c_1 \right) - \frac{9}{2}\left( \frac{2x + 3}{4}\sqrt{x^2 + 3x - 18} - \frac{81}{8}\text{ log }\left| \left( x + \frac{3}{2} \right) + \sqrt{x^2 + 3x - 18} \right| + c_2 \right)\]
\[ = \frac{1}{3} \left( x^2 + 3x - 18 \right)^\frac{3}{2} - \frac{9}{8}\left( 2x + 3 \right)\sqrt{x^2 + 3x - 18} + \frac{729}{16}\text{ log}\left| \left( x + \frac{3}{2} \right) + \sqrt{x^2 + 3x - 18} \right| + c\]
\[\text{ Hence,} \int\left( x - 3 \right)\sqrt{x^2 + 3x - 18} \text{  dx }= \frac{1}{3} \left( x^2 + 3x - 18 \right)^\frac{3}{2} - \frac{9}{8}\left( 2x + 3 \right)\sqrt{x^2 + 3x - 18} + \frac{729}{16}\log\left| \left( x + \frac{3}{2} \right) + \sqrt{x^2 + 3x - 18} \right| + c\]