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How to solve this:

[shyamjayakannan]

[tex]\displaystyle\int\limits_{1}^{2}x\log_2\frac{1}{x}\ dx=\int\limits_{1}^{2}x\frac{\displaystyle\ln\frac{1}{x}}{\displaystyle\ln2}\ dx=\frac{1}{\ln2}\int\limits_{1}^{2}x\ln\frac{1}{x}\ dx[/tex]. Let [tex]\ln\frac{1}{x}=t\Rightarrow\frac{dt}{dx}=-\frac{1}{x}=-e^t[/tex]. Substituting in the integral with new limits, we get:

[tex]=-\frac{1}{\ln2}\int\limits_{0}^{\ln\frac{1}{2}}te^{-2t}\ dt=-\frac{1}{\ln2}\int\limits_{0}^{-\ln2}te^{-2t}\ dt[/tex]. Now, use integration by parts:

[tex]=-\frac{1}{\ln2}\left(\left.-t\frac{e^{-2t}}{2}\right|^{-\ln2}_0-\int\limits^{-\ln2}_{0}\left.-\frac{e^{-2t}}{2}\right|^{-\ln2}_0\ dt\right)=-\frac{1}{\ln2}\left(2\ln2-\left.\frac{e^{-2t}}{4}\right|^{-\ln2}_0\right)=-\frac{1}{\ln2}\left(2\ln2-\frac{3}{4}\right)=\boxed{\frac{3}{4\ln2}-2}[/tex]