calculus

(ln(x))^2 dx, integrate.

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  1. use integration by parts. That is just the product rule in reverse.

    d(uv) = u dv + v du
    u dv = d(uv) - v du
    ∫u dv = ∫d(uv) - ∫v du
    ∫u dv = uv - ∫v du

    So, here we just let

    u = (lnx)^2
    dv = dx

    du = 2lnx * 1/x dx = (2lnx)/x
    v = x

    ∫(lnx)^2 dx = x(lnx)^2 - ∫2lnx dx

    Now repeat, this time letting

    u = lnx
    dv = 2dx

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