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Question
Evaluate the following.
`int "x"^2 *"e"^"3x"`dx
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Solution
Let I = `int "x"^2 "e"^"3x"`dx
`= "x"^2 int "e"^"3x" "dx" - int["d"/"dx" ("x"^2) int "e"^"3x" "dx"]` dx
`= "x"^2 * ("e"^"3x"/3) - int 2"x" * "e"^"3x"/3` dx
`= ("x"^2)/3 "e"^"3x" - 2/3 int "x" * "e"^"3x"` dx
`= ("x"^2)/3 "e"^"3x" - 2/3 ["x" int "e"^"3x" "dx" - int ("d"/"dx" ("x") int "e"^"3x" "dx") "dx"]`
`= ("x"^2 * "e"^"3x")/3 - 2/3 ["x" * "e"^"3x"/3 - int 1 * "e"^"3x"/3 "dx"]`
`= ("x"^2 * "e"^"3x")/3 - 2/3 [1/3 "xe"^"3x" - 1/3 int "e"^"3x" "dx"]`
`= ("x"^2 * "e"^"3x")/3 - 2/3 [1/3 "xe"^"3x" - 1/3 * "e"^"3x"/3]` + c
∴ I = `1/3 "x"^2 * "e"^"3x" - 2/9 "xe"^"3x" + 2/27 "e"^"3x" + "c"`
Notes
The answer in the textbook is incorrect.
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