Question

Integrate ln (1/x) dx =?

Answer 1

`-xlnx+x+C`

Explanation:

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

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

Using integration by parts:

`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

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

Explanation:

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"`