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

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

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

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

$dx=e^tdt$

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