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

Question

20392 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-11T19:56:23

$1/lnx+C$

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 $u=lnx$ ,

$=1/lnx+C$

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