What is #int ((1+lnx)/x)^2dx#?

1 Answer
May 24, 2018

Answer:

The answer is #=-5/x-4ln(|x|)/x-(ln(|x|))^2/x+C#

Explanation:

The integral is

#I=int((1+lnx)^2dx)/x^2#

#=int((1+2lnx+(lnx)^2)dx)/x^2#

#=intdx/x^2+int(2lnxdx)/x^2+int((lnx)^2dx)/x^2#

#=I_1+I_2+I_3#

Therefore,

#I_1=intdx/x^2=intx^-2dx=-1/x#

For the calculation of #I_2#, perform an integration by parts

#intuv'=uv-intu'v#

#u=lnx#, #=>#, #u'=1/x#

#v'=2/x^2#, #=>#, #v=-2/x#

So,

#I_2=-2lnx/x-2int(-dx)/x^2#

#=-2lnx/x-2/x#

For the calculation of #I_3#, perform an integration by parts

#u=(lnx)^2#, #=>#, #u'=2lnx/x#

#v'=1/x^2#, #=>#, #v=-1/x#

#I_3=-(lnx)^2/x-2int(-lnxdx)/x^2#

#=-(lnx)^2/x-2lnx/x-2/x#

Finally,

#I=I_1+I_2+I_3#

#=-1/x-2lnx/x-2/x-(lnx)^2/x-2lnx/x-2/x+C#

#=-5/x-4lnx/x-(lnx)^2/x+C#