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

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.

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  • $\begingroup$ Try $u=\ln(x)$ and then using the series expansion. The integral is from 0 to 1 so it is valid. $\endgroup$ – Moya Jul 7 '19 at 19:21
  • $\begingroup$ $u=ln(x)+1$ 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 :) $\endgroup$ – thomasfermi Jul 7 '19 at 19:41
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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 $$\int_1^e \frac{\ln x}{(1+\ln x)^2} dx = \left [ \frac{x}{1+ \ln x} \right]^e_1=\frac e2 -1$$

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  • $\begingroup$ Thanks, that was pretty big brain $\endgroup$ – Anirudh Nov 14 '19 at 6:44
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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*}

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    $\begingroup$ In your first integral the limits should be $0$ and $1$. $\endgroup$ – Anurag A Jul 7 '19 at 19:31
  • $\begingroup$ @AnuragA Thanks! I just fixed it $\endgroup$ – boink Jul 7 '19 at 19:41
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    $\begingroup$ I believe that you might have a sign error. Wolfram Alpha and my calculation gives $e/2-1$ as a result. $\endgroup$ – thomasfermi Jul 7 '19 at 19:44
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    $\begingroup$ @thomasfermi In addition to you and Wolfram Alpha, the positivity of the integrand on the interval of integration says that the answer should be positive. $\endgroup$ – Andreas Blass Jul 7 '19 at 19:47
  • $\begingroup$ Thanks—I swapped signs when integrating by parts apparently! $\endgroup$ – boink Jul 7 '19 at 19:48

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