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Question
Integrate the following with respect to x.
ex(1 + x) log(xex)
Solution
ex(1 + x) log(xex) = (ex + xex) log(xex)
Let z = xex
Then dz = d(xex)
dz = (xex + ex) dx .......(Using product rule)
So `int "e"^x (1 + x) log(x"e"^x) "d"x`
= `int log (x"e"^x) ("e"^x + x"e"^x) "d"x`
= `int log z "d"z`
= z(log z – 1) + c
= xex [log (xex) – 1] + c
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