How do you integrate #(ln x)^2 dx#?

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Answer 1 · 2016-05-22T00:27:49

Use integration by parts twice to find that

#intln^2(x)dx=xln^2(x)-2xln(x)+2x+C#

We will proceed using integration by parts :

Integration by parts (i):

Let #u = ln^2(x)# and #dv = dx#
Then #du = 2ln(x)/xdx# and #v = x#

Applying the integration by parts formula #intudv = uv - intvdu#

#intln^2(x)dx = xln^2(x) - int(2xln(x))/xdx#

#=xln^2(x)-2intln(x)dx" (1)"#

Integration by parts (ii):

Focusing on the remaining integral, let #u = ln(x)# and #dv = dx#
Then #du = 1/xdx# and #v = x#

Applying the formula:

#intln(x)dx = xln(x) - intx/xdx#

#=xln(x) - intdx#

#=xln(x) - x + C#

Substituting this back into #"(1)"#:

#intln^2(x)dx = xln^2(x)-2(xln(x)-x)+C#

#=xln^2(x)-2xln(x)+2x+C#