# Integrals: $I=\int_{1}^{e}\frac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$

Evaluate

$$I=\int_{1}^{e}\dfrac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$$

Thank you very much

-
You're very welcome. And your question is...? –  DonAntonio Dec 27 '12 at 0:45
You mean "Evaluate $I=\cdots$"? –  Pedro Tamaroff Dec 27 '12 at 0:48
I suppose this is homework of some kind, but even if it isn't, it would be nice to see your ideas. –  user50407 Dec 27 '12 at 1:07
Yes, the question as it is shows no effort on the part of the questioner. –  Ben Millwood Dec 27 '12 at 1:22

First make a change of variable to get the integral $$I = \int_0^1 \frac{\mathrm{e}^y y \left( y + 1 \right)}{\left( \mathrm{e}^y + y + 1 \right)^3} \mathrm{d} y$$ and notice that the integrand is obtained as a derivative of $$- \tfrac{\left( y + 1 \right)^2}{2 \left( \mathrm{e}^y + y + 1 \right)^2}$$ Now apply the fundamental theorem of calculus to get $$I = \frac{1}{8} - \frac{2}{\left( 2 + \mathrm{e} \right)^2}$$

-
The "and notice that..." and etc. part doesn't seem to be straightforward. Most lecturers/instructors would require to see how was that achieved. –  DonAntonio Dec 27 '12 at 2:43
@DonAntonio Is it good policy to these homework questions to give detailed and complete answers? (especially where there were no efforts in the question?) –  Learner Dec 27 '12 at 2:52
Well, you did give the complete final answer, yet you skipped an intermediate step...Anyway, I think the OP would profit from seeing how did you get that function's primitive. –  DonAntonio Dec 27 '12 at 2:58

$$I=\int\limits_{1}^{e}\frac{\ln x(\ln x+1)}{{x}^{3}(\frac{1}{x}+1+\frac{\ln x}{x})^3}\,\mathrm dx$$

Set: $$t=\frac{1}{x}+1+\frac{\ln x}{x}$$

So $t-1=\frac{\ln x+1}{x}$

And $$\mathrm dt=-\frac{\ln x}{x^2}\mathrm dx$$

When $x=1$, $t=2$

When $x=e$, $t=\frac{2+e}{e}$

So $$I=-\int_{2}^{\frac{2+e}{e}}\frac{t-1}{t^3}\,dt$$

-
$$I=\int_{1}^{e}\frac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$$
Let, $\ln{x}=t$, then, $x=e^{t},\ln x=t$
$dx=e^tdt$
$$I=\int_{0}^{1}\frac{e^t(t)(t+1)}{(1+e^t+t)^3}dt$$