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Question
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
Sum
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Solution
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right)\left( 1 - 2x \right)dx\]
\[ = \int\left( 2 - 3x \right) \left( 3 - 6x + 2x - 4 x^2 \right)dx\]
\[ = \int\left( 2 - 3x \right) \left( - 4 x^2 - 4x + 3 \right)dx\]
\[ = \int\left( - 8 x^2 - 8x + 6 + 12 x^3 + 12 x^2 - 9x \right)dx\]
\[ = \int\left( 12 x^3 + 4 x^2 - 17x + 6 \right)dx\]
\[ = \frac{12 x^4}{4} + \frac{4 x^3}{3} - \frac{17 x^2}{2} + 6x + C\]
\[ = 3 x^4 + \frac{4}{3} x^3 - \frac{17}{2} x^2 + 6x + C\]
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Find: `int (3x +5)/(x^2+3x-18)dx.`
