int {{x^x}(1 + log x),dx} is equal to | vedclass

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Question

(A) Let $I = \int {{x^x}(1 + \log x)\,dx} $. Consider the function $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\log f(x) = x \log

Vedclass · Accepted Answer

${x^x} + C$

Vedclass · Answer

(A) Let $I = \int {{x^x}(1 + \log x)\,dx} $. Consider the function $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\log f(x) = x \log x$. Differentiating both sides with respect to $x$,we get $\frac{1}{f(x)} f'(x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$f'(x) = f(x)(1 + \log x) = x^x(1 + \log x)$. Since the derivative of $x^x$ is $x^x(1 + \log x)$,the integral of $x^x(1 + \log x)$ is $x^x + C$.

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

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