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Q.
$\int \frac{ e ^{ x }( x -1)( x -\ln x )}{ x ^2} dx$ is equal to
(where $c$ is constant of integration)
Integrals
Solution:
$\int \frac{ e ^{ x }( x -1)( x -\ln x )}{ x ^2} dx =\int e ^{ x }\left(\frac{1}{ x }-\frac{1}{ x ^2}\right) \ln \left(\frac{ e ^{ x }}{ x }\right) dx$
Put $ t=\frac{e^x}{x}$
$=\int \ln t dt =( t \ln t - t )+ c =\frac{ e ^{ x }}{ x }\left(\ln \left(\frac{ e ^{ x }}{ x }\right)-1\right)+ c$
$=\frac{ e ^{ x }( x -\ln x -1)}{ x }+ c $