∫xx1+loge⁡xdx is equal to

$\int x^{x} (1 + \log_{e} x) d x$ is equal to

A
$x^{x} \log_{e} x + C$
B
$e^{x^{2}} + C$
C
$x^{x} + C$
D
none of these

Solution:

We have,

$\int x^{x} (1 + \log_{e} x) d x = \int 1 \cdot d (x^{x}) = x^{x} + C$

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