The value of the following definite integral is ______ (round off to three decimal places)

$$\int_1^e (x \: \ln \: x) dx$$

edited

$I = \int \underset{2^{nd}}{\underbrace{x}}\; \underset{1^{st}}{\underbrace{lnx}}\;dx$

$\Rightarrow I = (lnx)*\left ( \int x\;dx \right ) \;- \int \left [ \left ( \frac{\mathrm{d} (lnx)}{\mathrm{d} x} \right )\left (\int x\;dx \right ) \right ]\;dx$

$\Rightarrow \frac{x^{2}lnx}{2} - \int \frac{x^{2}}{2x}\;dx$

$\Rightarrow \frac{x^{2}lnx}{2} - \frac{x^{2}}{4} + c$ , where ‘c’ is an arbitrary constant.

Now, After putting limits, It becomes :-

$\frac{e^{2}}{2} - \frac{e^{2}}{4} - 0 + \frac{1}{4} =$  $\frac{(e^{2}+1)}{4} =$  $\frac{((2.718)^{2}+1)}{4} = 2.096$
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