Putting xe^x = t (xe^x + e^x) dx = dt ∫log t dt From by parts, we have t log t t + C xe^x log xe^x xe^x + C ...

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Integration by Parts

∫ x+1exlog xe...

Question

$\int (x + 1) e^{x} l o g (x e^{x}) d x$ .

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Solution

Putting $x e^{x} = t$
$(x e^{x} + e^{x}) d x = d t$
$\int log t d t$
From by parts, we have
$t log t - t + C$
$x e^{x} log x e^{x} - x e^{x} + C$

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