Question

Integrate the following with respect to x:

x log x

Answer 1

`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"`