How do you evaluate the integral $int x(lnx)^2dx$ ?

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Answer 1 · 2017-05-23T23:09:51

$intudv = uv - intvdu$

Let $u = (ln(x))^2 and dv = xdx$ , then $du =(2ln(x))/xdx and v = x^2/2$

$intx(ln(x))^2dx = (xln(x))^2/2 - intxln(x)dx$

Integration by parts for the second integral:

Let $u = ln(x) and dv = xdx$ , then $du =1/xdx and v = x^2/2$

$intx(ln(x))^2dx = (xln(x))^2/2 - x^2/2ln(x)+1/2intxdx$

$intx(ln(x))^2dx = (xln(x))^2/2 - x^2/2ln(x)+1/4x^2+ C$

How do you evaluate the integral int x(lnx)^2dxint x(lnx)^2dx?