Integrate ln (1/x) dx =?

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Answer 1 · 2018-04-26T15:28:03

#-xlnx+x+C#

Given: #intln(1/x) \ dx#

Notice how it equals #intln(1/x)*1 \ dx#.

#intu \ dv=uv-intv \ du#

Let #u=ln(1/x),:.du=-1/x \ dx# by the chain rule.

Then, #dv=1 \ dx, v=int1 \ dx=x#. We don't put the constant until we finish the whole integration.

Inputting, we get,

#intln(1/x) \ dx=xln(1/x)-intx*(-1/x) \ dx#

#=xln(1/x)-int-1 \ dx#

#=xlnx(1/x)-(-x)#

#=xln(1/x)+x#

We now simplify the #xln(1/x)# part.

Notice that #ln(1/x)=ln(x^-1)=-1lnx=-lnx# by the power rule for logarithms.

So we get:

#=x*-lnx+x#

#=-xlnx+x#

Finally, we add a constant, and so the final answer is:

#color(blue)(=barul|-xlnx+x+C|#

Answer 2 · 2018-04-26T16:39:01

#intln(1/x)dx=x(1-lnx)+"c"#

To find #intln(1"/"x)dx#, we use the integral of inverse functions theorem.

Let #g# be the inverse of a continuous function #f#. Let #F# be an antiderivative of #f#. Then

#intg(x)dx=xg(x)-F(g(x))+"c"#

Now, the inverse of #ln(1"/"x)# is #e^-x#. (I will leave it up to you to check that it is.) An antiderivative of #e^-x# is #-e^-x#. So

#intln(1/x)dx=xln(1/x)-(-e^(-ln(1/x)))+"c"=xln(x^-1)+x+"c"=x-xlnx+"c"=x(1-lnx)+"c"#