How do you evaluate the integral #int lnx/(1+x)^2#?

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Answer 1 · 2017-01-19T20:47:08

#-lnx/(1+x)+lnx-ln(1+x)+C#

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

Integration by parts takes the form #intudv=uv-intvdu#. Let:

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

Differentiate #u# and integrate #dv#. The integration of #dv# is best performed with the substitution #t=1+x=>dt=dx#. You should get:

#du=1/xdx#
#v=-1/(1+x)#

Then:

#I=uv-intvdu#

#I=-lnx/(1+x)-int1/x(-1/(1+x))dx#

#I=-lnx/(1+x)+int1/(x(1+x))dx#

Perform partial fraction decomposition on #1/(x(1+x))#:

#1/(x(1+x))=A/x+B/(1+x)#

Then:

#1=A(1+x)+Bx#

Letting #x=-1#:

#1=A(1-1)+B(-1)#

#B=-1#

Letting #x=0#:

#1=A(1+0)+B(0)#

#1=A#

Then:

#1/(x(1+x))=1/x-1/(1+x)#

So:

#I=-lnx/(1+x)+int1/xdx-int1/(1+x)dx#

These are simple integrals. The second can be performed with the substitution #s=1+x=>ds=dx#.

#I=-lnx/(1+x)+lnx-ln(1+x)+C#