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

x^{x} log x + k

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e^{x^{x}} + k

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x^{x} + k

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None of these

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Solution

The correct option is B $x^{x} + k$
Let $I = \int x^{x} (1 + log x) d x$
Substitute $t = x^{2} + 1 \Rightarrow d t = x^{x} (1 + log x) d x$
$∴ I = \int d t = t + k = x^{x} + k$

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