Ta có các đẳng thức sau:
$\dfrac{\ln (x)(\ln (x)+1)}{(1+x+\ln (x))^3}=
-{\frac {\ln\left( x \right) +1}{x \left( 1+x+\ln\left( x \right)
\right) ^{2}}}+{\frac { \left( \ln\left( x \right) +1 \right) ^{2}
\left( 1+\dfrac{1}{x} \right) }{ \left( 1+x+\ln\left( x \right)
\right) ^{3}}}$
$\int {\frac {\ln\left( x \right) +1}{x \left( 1+x+\ln\left( x \right)
\right) ^{2}}} dx={\frac {x}{ \left( 1+x \right)\left( 1+x+\ln\left( x \right)
\right) }}+\int {\frac {x+1+{x}^{2}}{ \left( 1+x+\ln\left( x
\right)\right) x \left( 1+x \right) ^{2}}}{dx}+C$
$\int {\frac { \left( \ln\left( x \right) +1 \right) ^{2} \left( 1+{x}^{-1
} \right) }{ \left( 1+x+\ln\left( x \right)\right) ^{3}}} dx
=\frac{1}{2}{\frac {x \left( 4+5x+4\ln\left( x \right) +{x}^{2}+2\ln
\left( x \right) x \right) }{ \left( 1+x \right)\left( 1+x+\ln
\left( x \right)\right) ^{2}}}+\int {\frac {x+1+{x}^{2}}{ \left(
1+x+\ln\left( x \right)\right) x \left( 1+x \right) ^{2}}}{dx}+C$
Suy ra $\int \dfrac{\ln (x)(\ln (x)+1)}{(1+x+\ln (x))^3}dx=-\frac{1}{2}{\frac { \left( \ln\left( x \right) +1 \right) ^{2}}{ \left( 1
+x+\ln\left( x \right)\right) ^{2}}}
+C
$
Suy ra $I=\int_{1}^{e}\dfrac{\ln (x)(\ln (x)+1)}{(1+x+\ln (x))^3}dx=\frac{1}{8}{\frac {-12+4{{\rm e}}+{{\rm e}^{2}}}{4+4{{\rm e}}+{
{\rm e}^{2}}}}
$
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