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Evaluate

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

Thank you very much

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    You're very welcome. And your question is...?2012-12-27
  • 0
    You mean "Evaluate $I=\cdots$"?2012-12-27
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    I suppose this is homework of some kind, but even if it isn't, it would be nice to see your ideas.2012-12-27
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    Yes, the question as it is shows no effort on the part of the questioner.2012-12-27

3 Answers 3

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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 $$ I = \frac{1}{8} - \frac{2}{\left( 2 + \mathrm{e} \right)^2} $$

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    The "and notice that..." and etc. part doesn't seem to be straightforward. Most lecturers/instructors would require to see how was that achieved.2012-12-27
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    @DonAntonio Is it good policy to these homework questions to give detailed and complete answers? (especially where there were no efforts in the question?)2012-12-27
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    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.2012-12-27
2

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

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

0

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

Let, $\ln{x}=t$, then, $x=e^{t},\ln x=t$

$dx=e^tdt$

$$I=\int_{0}^{1}\frac{e^t(t)(t+1)}{(1+e^t+t)^3}dt$$