Question

How do you integrate `-1 / (x(ln x)^2)`?

Answer 1

`1/lnx+C`

Explanation:

We have:

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

We can use substitution here, since the derivative of `lnx`, which is `1/x`, is present alongside `lnx`.

Let `u=lnx` such that `du=1/xdx`.

We then have:

`-int(lnx)^-2/xdx=-int(lnx)^-2(1/x)dx=-intu^-2du`

Integrate this with the rule: `intu^ndu=u^(n+1)/(n+1)+C`, where `n!=1`.

Thus,

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

Since `u=lnx`,

`=1/lnx+C`