Evaluate
$$I=\int_{1}^{e}\dfrac{\ln x(\ln x+1)}{(1+x+\ln x)^3}dx$$
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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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$$I=\int\limits_{1}^{e}\frac{\ln x(\ln x+1)}{{x}^{3}(\frac{1}{x}+1+\frac{\ln x}{x})^3}\,\mathrm dx$$ 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, $x=e^{t},\ln x=t$ $dx=e^tdt$ $$I=\int_{0}^{1}\frac{e^t(t)(t+1)}{(1+e^t+t)^3}dt$$ |
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