What is #int ln(2x)^2dx#?

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Answer 1 · 2018-07-28T17:20:16

#=2x\ln(2 x) -2x +C#

Given that

#\int ln(2x)^2 \ \d x#

#=\int 2ln(2x) \ \d x#

#=2\int (lnx+\ln 2) \ \d x#

#=2\int lnx\ dx+2\int \ln 2 \d x#

#=2(\ln x \int 1\ dx-\int (1/dx(lnx)\cdot \int1 \ dx) dx)+2\ln 2\int 1 \d x#

#=2(x\ln x -\int (1/x\cdot x) dx)+2\ln 2 \ (x)+C#

#=2(x\ln x -x)+2x\ln 2 +C#

#=2x\ln(2 x) -2x +C#

Answer 2 · 2018-07-28T19:05:02

The answer is #=2xln2x-2x+C#

A slightly different method

The integral is

#I=intln(2x)^2dx=2intln2xdx#

#intuv'dx=uv-intu'vdx#

#u=ln2x#, #=>#, #u'=1/2x*2=1/x#

#v'=1#, #=>#, #v=x#

Therefore,

#I=2xln2x-2int1/x*xdx#

#=2xln2x-2intdx#

#=2xln2x-2x+C#