Posted by lucy on Saturday, December 29, 2012 at 8:09am.
integrate by parts
x(lnx)^2
∫u dv = uv - ∫v du
So in ∫x(lnx)^2 dx
let u = (lnx)^2 and dv = x dx
du/dx = 2lnx (1/x) and v = (1/2)x^2
du = (2/x)(lnx) dx and v = (1/2)x^2
∫x(lnx)^2 dx = (1/2)(x^2)(lnx)^2 - ∫(1/2)x^2 (2/x)(lnx) dx
= (1/2)(x^2)(lnx)^2 - ∫(x)(lnx) dx
now let's do ∫(x)(lnx) dx
let u = lnx and dv = x dx
du/dx = 1/x and v = (1/2)x^2
du = (1/x) dx and v = (1/2)x^2
∫(x)(lnx) dx = (lnx)(1/2)(x^2) - ∫(1/2)(x^2)(1/x) dx
= (1/2)(x^2)(lnx) - ∫(1/2)x dx
= (1/2)(x^2)(lnx) - (1/4)x^2
So finally ...
∫x(lnx)^2 dx
= (1/2)(x^2)(lnx)^2 - ∫(x)(lnx) dx
= (1/2)(x^2)(lnx)^2 - ((1/2)(x^2)(lnx) - (1/4)x^2) )
= (1/2)(x^2)(lnx)^2 - (1/2)(x^2)(lnx) + (1/4)x^2)
better check my algebra, I should have written it out on paper first.
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