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

1 Answer
Jun 24, 2018

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