How can I integrate $\frac{\ln x - 1}{(\ln x)^2}$?

Question

I've been stuggling with integrating this: $$ \frac{\ln x - 1}{(\ln x)^2} $$ Could you help? My guess is integration by parts but can't figure out how. Thanks!

Answer 1

Setting $\ln x=y\iff x=e^y$

$$\int\frac{\ln x-1}{(\ln x)^2}dx=\int\frac{y-1}{y^2}e^y\ dy=\int\left[\frac1y-\frac1{y^2}\right]e^ydy$$

Now, $\displaystyle\int e^y[f(y)+f'(y)]dy= e^yf(y)$

---

Alternatively integrating by Parts, $$\int\frac{dx}{\ln x}=\int 1\cdot\frac1{\ln x}dx$$

$$=\frac1{\ln x}\int dx-\int\left(\frac{d(1/\ln x)}{dx}\int dx\right)dx$$

$$=\cdots$$ $$=\frac x{\ln x}+\int\frac{dx}{(\ln x)^2}$$

Answer 2

Another way to do it by parts: Let

$$u=x(\ln x-1)\qquad\text{and}\qquad dv={dx\over x(\ln x)^2}$$

so that

$$du=\ln x\,dx\qquad\text{and}\qquad v={-1\over\ln x}$$

The key is inserting an $x$ in with the $1/(\ln x)^2$ to make it the derivative of $-1/\ln x$. What makes it work, though, is that the $du$ turns out just right.