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Question
Integrate the following with respect to x:
25xe–5x
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Solution
`int 5x"e"^(-5x) "d"x = int 5x"e"^(- 5x) "d"x`
`int "u" "dv" = "uv" - "u""'""v"_2 + "u""'""v"_2 - "u""'""v"_3 +` .........(1)
u = x
u' = 1
u" = 0
dv = `"e"^(- 5x) * "d"x`
⇒ v = `int "e"^(- 5x) * d"x`
= `("e"^(- 5x))/(- 5)`
v1 = `int "v" "d"x`
= `- 1/5 int "e"^(-5x) * "d"x`
= `- 1/5 xx ("e"^(- 5x))/(- 5)`
= `+ 1/5^2 "e"^(-5x)`
v2 = `int "v"_1 "d"x`
= `int 1/5^2 "e"^(- 5x) "d"x`
= `1/5^2 xx ("e"^-5x)/(- 5)`
= `- 1/5^3 "e"^(- 5x)`
`int 25x "e"^(- 5x) "d"x = 25[x xx ("e"^(- 5x))/(5x) - 1 xx 1/5^2 "e"^(- 5x) + 0 xx - 1/5^3 "e"^(-5x)]`
= `25[ - x/5 "e"^(- 5x) - 1/5^2 "e"(- 5x)]`
= `- 5x "e"^(-5 x) - "e"^(- 5x)`
`int 25x "e"^(- 5x) "d"x = - "e"^( 5x) (5x + 1 + "c"`
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