How to evaluate the integral $\int_{1}^{e} \frac{\ln(x)}{\left( 1+\ln(x) \right)^2}dx$?

Question

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

I consider using the u-subs $u=\ln(x)$ or $u=\ln(x)+1$, but I was left with a factor of $e^u$ I couldn't get rid of.

I was also considering usign the series expansion of $\frac{1}{(1-x)^2}$, but I couldn't find a way to easily integrate powers of $\ln(x)$. Any insight or help would be greatly appreciated.

Answer 1

The square in the denominator should remind you of the quotient rule. looking at the integrand and rewriting it as the derivative of $\frac{u}{\ln x +1}$with the quotient rule you get $$\frac{\ln x}{(1+\ln x)^2}=\frac{u'(\ln x +1)-\frac ux}{(1+\ln x)^2}$$ from which you can guess that $u=x$. This indicates that $$\int_1^e \frac{\ln x}{(1+\ln x)^2} dx = \left [ \frac{x}{1+ \ln x} \right]^e_1=\frac e2 -1$$

Answer 2

Substituting $t=\ln x$ gives us that the integral $I$ satisfies $$I=\int_0^1\frac{e^tt}{(t+1)^2}dt.$$ Now we integrate by parts with $u=e^tt$ and $dv=\frac{1}{(t+1)^2}dt$. In particular, we get that $du=e^tt+e^t$ and $v=-\frac{1}{t+1}$. This implies that\begin{align*}I&=\int_0^1\frac{e^t(t+1)}{t+1}dt-\frac{e^tt}{t+1}\Bigg|_0^1\\&=-1+e-\frac e2=\boxed{\frac e2-1.}\end{align*}