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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread
![](https://latex.codecogs.com/png.latex?\bg_white \int_1^e x\ln{x} dx = \frac{1}{2}\int_1^e (2x\ln{x} + x - x)dx = \frac{1}{2} \left ( x^2 \ln{x} - \frac{x^2}{2} \right )_1^e)
![](https://latex.codecogs.com/png.latex?\bg_white = \frac{1}{2}\left ( e^2\ln{e} - \frac{e^2}{2} - 1^2\ln{1} + \frac{1^2}{2} \right ) = \frac{e^2 +1}{4})
![](https://latex.codecogs.com/png.latex?\bg_white $Wolfram Alpha verifies this result. The answer given is simply incorrect.$)
Find d/dx ( x^2logx) ... = 2xlnx + x
Hence evaluate Integral from 1 to e xlnx dx
I get the answer : (e^2 + 1)/4
Could someone see if they get the same result.
The answer says its (e^2 - e + 1)/ 2
Still can't figure where I have gone wrong.