Question

What is `int1/(xln^3x)dx` ?

`int1/(xln^3x)dx`

Answer 1

`int 1/(xln^3x)dx =-1/(2ln^2x) + C`

Explanation:

We will use integration by substitution:

Let `u = lnx`. Then `(du)/dx = 1/x => dx=xdu`.

Thus,

`int 1/(xln^3x) dx = int 1/(xu^3) * xdu = int 1/u^3 du`

As you can see, the `x` cancells nicely.

We have to apply the power rule now:

`int u^n du = u^(n+1)/(n+1) + C` with `n = -3`.

`int 1/u^3 du = u^(-2)/-2 +C= -1/(2u^2)+C`

Undo the substitution :

`= -1/(2ln^2x) + C`

`color(red)( :. int 1/(xln^3x) dx =-1/(2ln^2x) + C`