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Question
Integrate the following with respect to x:
9xe3x
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Solution
`int 9x"e"^(3x) "d"x = 9 int x"e"^(3x) "d"x`
`int "u" "dv" = "uv" - "u'v"_1 + "u''v"_2 - "u""'''""v"_3 + "u""''""v"_4 -` ...........(1)
u = x
u' = 1
u = 0
dv = `"e"^(3x) * "d"x`
v = `int "e"^(3x) * "d"x`
= `("e"^(3x))/3`
v1 = `int "v" "d"x`
= `int ("e"^(3x))/3 * "d"x`
= `1/3 xx 1/3 "e"^(3x)`
v2 = `int "v"_1 "d"x`
= `int 1/3^2 "e"^(3x) * "d"x`
= `1/3^2 xx 1/3 "e"^(3x)`
`9int x "e"^(3x) "d"x = 9[x ("e"^(3x))/3 - 1 xx 1/3^2 "e"^(3x) + 0 xx 1/3^2 "e"^(3x)]`
`9int x "e"^(3x) "d"x = 9[x/3 "e"^(3x) - 1/3^2 "e"^(3x)]`
`9int x "e"^(3x) "d"x = (3x - 1) "e"^(3x) + "c"`
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