Question

What is the indefinite integral of `-1 / (x(ln x)^2)`?

Answer 1

`int-1/(x(lnx)^2)dx=1/lnx+C`

Explanation:

We first begin by taking that negative sign out of the integral:

`-int1/(x(lnx)^2)dx`

If we think about this, we see that we have a function, `lnx`, and its derivative, `1/x`, in the same integral. Don't be bothered by the fact that `lnx` is in the denominator; all that matters is this integral can be solved with a `u`-substitution:
Let `color(red)(u)=color(red)lnx->(du)/dx=1/x->color(blue)(du)=color(blue)(1/xdx)`

Applying this substitution to `-intcolor(blue)(1/x)1/(color(red)(lnx)^2)color(blue)(dx)`, we have:
`-int1/(u^2)du`

This is equivalent to `-intu^(-2)du`, which can be solved using the reverse power rule:
`-intu^(-2)du=-(-u^(-1))=1/u+C`

Because `u=lnx`, we can back-substitute to end up with:
`1/lnx+C`