Integral Ln|x|/(1+x)^2 dx Answer Guys ???

1 Answer

Explanation:

Let

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

Apply integration by parts:

#u(x)=ln|x|#, #u'(x)=1/x#.
#v'(x)=1/(1+x)^2#, #v(x)=-1/(1+x)#.

Hence

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

Apply partial fraction decomposition:

#I=-(ln|x|)/(1+x)+int(1/x-1/(1+x))dx#

Integrate term by term:

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

Simplify:

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