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

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

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.

• Try $u=\ln(x)$ and then using the series expansion. The integral is from 0 to 1 so it is valid. – Moya Jul 7 '19 at 19:21
• $u=ln(x)+1$ works nicely. The resulting integrand is a sum ($e^u/u$ and $-e^u/u^2$). Apply integration by parts to the $e^u/u^2$ integral and you will see some nice cancellation in the sum :) – thomasfermi Jul 7 '19 at 19:41

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 $$\int_1^e \frac{\ln x}{(1+\ln x)^2} dx = \left [ \frac{x}{1+ \ln x} \right]^e_1=\frac e2 -1$$
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*}
• In your first integral the limits should be $0$ and $1$. – Anurag A Jul 7 '19 at 19:31
• I believe that you might have a sign error. Wolfram Alpha and my calculation gives $e/2-1$ as a result. – thomasfermi Jul 7 '19 at 19:44