Implicit Differentiation ==> Integration Our class finished integration, like taking the integral of $5x^2$ and $\sin{x}$. 
For an extra credit problem, we were assigned:
1a. $\int x^{x^x}x^x\left(\ln^2(x)+\ln{x}+\frac1x\right) \,dx$
I had no clue how to do this, so I randomly took the derivative of $x^{x^x}$ using implicit differentiation and happened to get this exact integrand. However, how can I do this WITHOUT guess and check.
I want to do this by myself, so as an actual question, how would I go about integrating
$∫x^x(1+\ln{x})\,dx$, which $= x^x +C$. 
 A: 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
