How do you find #\int \ln x d x#?

Question

766 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-04T20:26:16

The answer is #=x(ln(x)-1)+C#

#intu'v=uv-intuv'#

Here, we have

#u'=1#, #=>#, #u=x#

#v=lnx#, #=>#, #v'=1/x#

Therefore,

#intlnxdx=xlnx-intx*1/xdx#

#=xlnx-int1dx#

#=xlnx-x+C#

#=x(ln(x)-1)+C#