integration

$\int _{ 1 }^{ 2 }{ { e }^{ ln\quad x }\cdot ln\quad x } \quad dx\\ =\int _{ 1 }^{ 2 }{ x\cdot ln\quad x } \quad dx\\ =\int _{ 1 }^{ 2 }{ \cfrac { { x }^{ 2 }(2\cdot ln\quad x\quad -1) }{ 4 } } \\ =(\cfrac { { 2 }^{ 2 }(2ln\quad 2\quad -1) }{ 4 } )-(\cfrac { 2ln\quad 1\quad -1 }{ 4 } )\\ =2ln\quad 2\quad -1-\cfrac { 1 }{ 4 } \\ \approx 0.636\\$

As we know, (x(e^x))`=(e^x)(1+x) By deducing this one to that form we get : (e^(ln(x))) ln(x) = (1/2)((e^(2ln(x))) (1+2ln(x))) - (1/2)((e^(2ln(x))) By integrating the RHS, We get: I = { [ln(x) (x^2)/2] - (x^2)/4 } between the limits 2 and 1 By further solving, we get 0.6362.

e^lnx=x ,so I changes to xlnx ,now integrate by parts to get the value .6346

One can also use Integration by Parts and then substitute the limits .