Integral Ln|x|/(1+x)^2 dx Answer Guys ???
1 Answer
Jun 24, 2018
Use integration by parts.
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#