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evaluate the integral (3+2x-x^2)^(3/2) dx
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\[\int\limits_{}^{} \sqrt[]{(3+2x-x^2)^3} dx\] Now, let's get rid the the square root by writing the inside as something squared. To do that we will need to complete the square and choose a trig sub such that the inside is just something square.
Do you know how to complete the square?
The inside is \[-x^2+2x+3\] \[=-(x^2-2x)+3\] \[=-(x^2-2x+?)+3+?\] What do we need to replace question with so that we can write that one part as something square. Like this: \[=-(x- \text{ blah} )^2+3+?\]
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