If $x^x = a$ then define $f(a) = x$.$f$ is the inverse of $x^x$. What is then f? And what is the derivative and antiderivative of f? What is the derivation for the answer?
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You can search for Lambert function – MattG88 Dec 22 '16 at 21:19
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See for example here. – Eff Dec 22 '16 at 21:20
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1One rewrite it as $(\ln x)e^{\ln x}=\ln a$ ($a>0$) then use Lambert's function. – Olivier Oloa Dec 22 '16 at 21:20
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The solution is given on the Wikipedia of the Lambert W function, which states that
$$x^x=a\implies x=\frac{\ln a}{W(\ln a)}=e^{W(\ln a)}$$
where $W(x)$ is the Lambert W function.
If you wish to take the derivative of $f(x)$, use the inverse function derivative rule:
$$f'(x)=\frac1{g'(f(x))}$$
where $g(x)$ is the inverse of $f(x)$. Thus,
$$g'(x)=g(x)(1+\ln x)\implies g'(f(x))=x\left(1+\ln\left(e^{W(\ln x)}\right)\right)=x(1+W(\ln x))$$
the $x$ in front is due to $f$ and $g$ being inverses.
$$f'(x)=\frac1{x(1+W(\ln x))}$$
To take the antiderivative, use the antiderivative of the inverse function formula.
$$\int f(x)\ dx=xf(x)-G(f(x))+c$$
where $G$ is the antiderivative of $g=x^x$. Unfortunately, there is no closed form for $G$. However, some information as to a series expansion and such is found in this question.
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