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$\int x^{x} \ln (e x) d x$ is equal to

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$x . \ln x + C$

$x^{\ln x} + C$

$x^{x} + C$

${(\ln x)}^{x} + C$

answer is A.

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Detailed Solution

$I = \int x^{x} \ln (e x) d x = \int x^{x} (1 + \ln x) d x$

Let $t = x^{x} = e^{x \ln x} \Rightarrow \frac{d t}{d x} = x^{x} (x . \frac{1}{x} + \ln x)$

$\Rightarrow d t = x^{x} (1 + \ln x) d x \Rightarrow I = \int d t = t + C = x^{x} + C$

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