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