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Question
Integrate the following with respect to x:
x log x
Solution
`int x log x "d"x`
u = x
u' = 1
u" = 0
dv = `log x "d"x`
⇒ v = `int log x "d"x`
u = `log x`
dv = dx
du = `1/x "d"x`
v = x
v = `x log x - int x xx 1/x "d"x`
v = `x log x - int "d"x`
⇒ v = `x log x - x`
v1 = `int "v" "d"x`
= `int (x log x - x) "d"x`
= `int x log x "d"x int x "d"x` .........(1)
u = x
dv = `log x "d"x`
d = dx
v = `int log x "d"x`
= `(x log x - x)`
`int log x "d"x = x(x log x - x) - int (x log x - x) "d"x`
`int log x "d"x = x^2 log x - x^2 - int x log x "d"x + int x "d"x`
`int x log x "d"x + int x log x "d"x = x^2 log x - xx^2 + x^2/2 2int x log x "d"x`
= `x^2 log x - x^2/2`
`int x log x "d"x = x^2/2 log x = x^2/4`
(1) ⇒ v1 = `x^2/2 log x - x^2/4 - int x "d"x`
= `x^2/2 log x - x^2/4 - x^2/2`
v1 = `x^2/2 log x - 3/4 x^2`
v2 = `int "v"_1 "d"x`
= `int (x^2/2 log x - 3/4 x^2) "d"x`
`int "u" "dv"` = uv – u'v1 + u"v2 – u"'v3 + ..........
`int x log x "d"x = x(x log x - x) - 1 (x^2/2 log x - 3/4 x^2) + 0 xx int "v"_1 "d"x`
`int x log x "d"x = x(x log x - x) = x^2/2 log x + 3/4 x^2 + "c"`
= `x^2 log x - x^2 - x^2/2 log x + 3/4 x^2 + "c"`
= `(x^2 - x^2/2) log x + 3/4 x^2 - x^2 + "c"`
= `9(2x^2 - x^2)/2) log x + (3/4 - 1) x^2 + "c"`
`(x^2/2) log x - 1/4 x^2 + "c"`
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