∫xx(1+log⁡x)dx is equal to

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

A
$x^{x}$ +c
B
$x^{2 x}$ +c
C
$x^{x} \log x$ +c
D
$1 / 2 (1 + \log x)^{2}$ +c

Solution:

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

Put $x^{x} = t$ , then $x^{x} (1 + \log x) d x = d t$

$∴ I = \int d t \Rightarrow I = t + C \Rightarrow I = x^{x} + C$

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