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

x^{x} + C

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x \cdot I n x + C

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(I n x)^{x} + C

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x^{I n x} + C

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Solution

The correct option is A $x^{x} + C$
$\int x^{x} l n e x d x$
$= \int x^{x} (l n e + l n x) d x$
$= \int x^{x} (1 + l n x) d x$
Let $x^{x} = t$
$x l n x = l n t$
$(1 + l n x) d x = \frac{1}{t} d t$
$= \int t \cdot \frac{1}{t} d t$
$= \int d t$
$= t + c$
$= x^{x} + c$
$∴ \int x^{x} l n e x d x = x^{x} + c$ .

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