Find $\int{\frac{1}{\left(1+\ln x\right)^2}\;dx}$

Question

How would you integrate a function almost entirely in logarithmic form, such as:$$\int{\frac{1}{\left(1+\ln x\right)^2}\;dx}$$

I have tried various substitutions and considered integrating by parts, but I can't seem to get a quick and elegant solution. What is the easiest way to do this?

Answer 1

Here's a clever trick:

1 is equal to $\ln e$, and $\log(a)+\log(b)=\log(ab)$.

So you could write your equation as

$$\int{\frac{1}{\left(\ln e+\ln x\right)^2}\;dx}$$

$$\int{\frac{1}{(\ln ex)^2}\;dx}$$

From here, we could substitute u for $ex$. The This makes $\frac{du}{dx} = e$, so $dx = \frac{du}{e}$

$$\frac{1}{e}\int{\frac{1}{(\ln u)^2}du}$$

I think you can take it froim here.

Answer 2

Let $u=1+\ln x$ ,

Then $x=e^{u-1}$

$dx=e^{u-1}~du$

$\therefore\int\dfrac{1}{(1+\ln x)^2}dx$

$=\int\dfrac{e^{u-1}}{u^2}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{u^{n-2}}{en!}du$

$=\int\left(\dfrac{1}{eu^2}+\dfrac{1}{eu}+\sum\limits_{n=2}^\infty\dfrac{u^{n-2}}{en!}\right)du$

$=-\dfrac{1}{eu}+\dfrac{\ln u}{e}+\sum\limits_{n=2}^\infty\dfrac{u^{n-1}}{en!(n-1)}+C$

$=-\dfrac{1}{e(1+\ln x)}+\dfrac{\ln(1+\ln x)}{e}+\sum\limits_{n=2}^\infty\dfrac{(1+\ln x)^{n-1}}{en!(n-1)}+C$