How do you integrate #int ln 4x dx # using integration by parts?

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Answer 1 · 2018-04-06T08:44:24

#=xln4x-x+C#

#intln4xdx#

You can't integrate #ln4x# so you pick

#u=ln4x# #rarr# #du=1/xdx#
#dv=dx##rarr##v=x#

And by using the formula:
#intudv=uv-intvdu#

#intln4xdx=xln4x-intx/xdx#

#=xln4x-int1dx#

#=xln4x-x+C#

Answer 2 · 2018-04-06T08:44:43

The answer is #=x(ln(|4x|)-1)+C#

#intuv'=uv-intu'v#

Here,

#u=ln4x#, #=>#, #u'=1/(4x)*4=1/x#

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

Therefore,

#intln4x=xln4x-int1/x*xdx#

#=xln4x-x+C#

#=x(ln(|4x|)-1)+C#