Integration of ; ((1/X(4 + Inx))dx?

1 Answer
Apr 29, 2018

#int1/(x(4+lnx))dx=ln|4+lnx|+C#

Explanation:

So, we want to determine

#int1/(x(4+lnx))dx#

This can be solved using a substitution:

#u=lnx#

#du=dx/x#

Rewriting the integral a little, we see that #du# indeed shows up:

#int1/(4+lnx)*dx/x#

So, apply the substitution:

#int(du)/(4+u)#

We can apply a second brief substitution here:

#v=4+u#

#dv=du#

#int(dv)/v=ln|v|+C#

#=ln|4+u|+C#

Rewrite in terms of #x:#

#int1/(x(4+lnx))dx=ln|4+lnx|+C#