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Question
Evaluate: `int "dx"/(3 - 2"x" - "x"^2)`
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Solution
Let I = `int "dx"/(3 - 2"x" - "x"^2)`
3 - 2x - x2 = - x2 - 2x + 3
= -(x2 + 2x - 3)
= - (x2 + 2x + 1 - 4)
= - [(x + 1)2 - 4]
= (2)2 - (x + 1)2
∴ I = `int "dx"/((2)^2 - ("x + 1")^2)`
`= 1/(2(2)) log |(2 + "x" + 1)/(2 - ("x + 1"))|` + c
∴ I = `1/4 log |(3 + "x")/(1 - "x")|` + c
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