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December 20, 2014

December 20, 2014

Posted by **lucy** on Saturday, December 29, 2012 at 8:09am.

x (lnx)^2 dx

- maths -
**lucy**, Saturday, December 29, 2012 at 8:10amintegrate by parts

x(lnx)^2

- maths -
**Reiny**, Saturday, December 29, 2012 at 8:46am∫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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