Notice that if you differentiate the equation:
$$f(x) = e^{2x} + 2xe^{2x} + e^{-x}*f(x)$$
$$f(x)(1 - e^{-x}) = e^{2x} + 2xe^{2x}$$
$$f(x) = \frac{e^{2x} + 2xe^{2x}}{1 - e^{-x}}$$
Start with $$\ln{x^{x^{x}}} = x^x \ln {x}$$
as this moves the power of $x$ down to become part of a product.
By differentiating both sides with respect to $x$ we may now use the product rule on one side:
$$\displaystyle\frac{1}{x^{x^x}}\displaystyle\frac{d}{dx}x^{x^x}=x^x \displaystyle\frac{1}{x} + \ln {x} \displaystyle\frac{d}{dx}x^x$$.
We can use the similar fact that
$$\ln{x^x} = x \ln {x}$$
and differentiate both sides again to get
$$\displaystyle\frac{1}{x^{x}}\displaystyle\frac{d}{dx}x^{x}=x \displaystyle\frac{1}{x} + \ln {x} \displaystyle\frac{d}{dx}x$$
which simplifies to
$$\displaystyle\frac{d}{dx}x^{x}= x^x(1 + \ln {x})$$.
Subtituting this into our first differentiation we have
$$\displaystyle\frac{1}{x^{x^x}}\displaystyle\frac{d}{dx}x^{x^x}=x^x \displaystyle\frac{1}{x} + x^x \ln {x} (1 + \ln {x})$$
which can then be rearranged to form:
$$\displaystyle\frac{d}{dx}x^{x^x}=x^{x^x} x^x (\ln^2{x}+\ln x + \displaystyle\frac{1}{x})$$
Side note Often the best ways to solve complicated integrals are:
a) Employing various techniques (integration by parts, substitution, recurrence relations etc.)
b) Rearranging the integrand to look more palatable/like a known derivative
c) Observing a pattern in the expression
Best Answer
The question is from a very long time ago, but I can still attempt to provide an answer. Implicit integration is kind of like the topic in differential equations called exact differential equations. It’s pretty much tracing backwards from applying multivariable chain rule on a function of multiple variables. I’d say it’s the closest thing I’ve seen to a concept of “implicit integration”.