∫ (e^x)lnx dx =?

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∫ (e^x)lnx dx =?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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use integration by parts
Laplace20, are u sure u can get the answer by using integration by parts?????

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im never sure about anything. why dont you try it
Integration by parts impossible to solve it. 100% impossible!!!
u = ln x , dv = e^x , du = 1/x , v = e^x = ln x * e^x - integral e^x *1/x
nevermind, this cant be integrated
this is not an elementary function
algebraic, log, inverse trig, trig, exponential,
what is this a trick question?
no, this can be done using integration by parts , just you need to apply it twice
its like how you integrate e^x sin(x) or e^(x)cos(x)
contining from above, ie I (integral we want ) = ln x * e^x - integral e^x *1/x
let u=e^x , dv = (1/x) dx
so du= e^x dx , and v= ln(x)
SO I = e^x ln(x) - [ e^(x) ln(x) - integral ( e^x ln(x) dx ) ]
but that integral e^x lnx is what we started with, it was "I" so I = I huh , thats a bit strange that method didnt work :|

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