Question

How do you find the indefinite integral of `int 1/(xlnx^3)`?

Answer 1

`int (dx)/(xlnx^3) = 1/3ln(ln x) +C`

Explanation:

We begin by exploiting the properties of logarithms and write:

`int (dx)/(xlnx^3) = int (dx)/(3xlnx)`

Now we note that `d(lnx) = (dx)/x`, so that:

`int (dx)/(3xlnx) = 1/3 int (d(lnx))/lnx= 1/3ln(ln x) +C`

Answer 2

Substitute `x=e^u` to get `1/3 ln(ln)x+C`

Explanation:

`I=int1/(x ln x^3)dx=1/3 int 1/(x ln x)dx` by fundamental law of logarithms.
Now substitute `x=e^u`, `ln x = u` and `dx=e^u du`
to get `I=1/3int1/(e^u u)e^udu=1/3 int1/u du=1/3ln x + C`.
(Or just substitute `x=e^u` at the start.)

Answer 3

You may have also meant to type `int1/(x(lnx)^3)dx`. If this is the case, look here; if not, you can still learn something from this!

If we do indeed have `int1/(x(lnx)^3)dx`, let `u=lnx`. This implies that `du=1/xdx`.

Then we have the equivalent integrals `int1/(lnx)^3(1/xdx)=int1/u^3du=intu^-3du`.

Now we can use `intu^ndu=u^(n+1)/(n+1)+C`, where `n!=-1`, which is not an issue here.

The integral then becomes `(lnx)^-2/(-2)+C`, or `(-1)/(2(lnx)^2)+C`.