A strange integral having to do with the sophomore's dream: I recently noticed that this really weird equation actually carries a closed form!
$$\int_0^1 \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x=0$$
I honestly do not know how to prove this amazing result! I do not know nearly enough about the sophomore's dream integral properties to answer this question, which I have been trying to apply here. (If possible, please stay with real methods, as I do not know contour integration yet)
 A: It is important to note some symmetry:
Consider: $$I=\int_0^1 \frac{x^x}{(1-x)^{1-x}}\text{d}x$$
If you substitute $x\to 1-x, \text{d}x\to -\text{d}x$
You are left with:
$$I=-\int_{1-0}^{1-1}\frac{(1-x)^{(1-x)}}{(1-(1-x))^{(1-(1-x))}}\text{d}x=\int_0^1 \frac{(1-x)^{1-x}}{x^x}\text{d}x$$
And your integral is only $I-I=0$, your result. You didn't even need to know $x^x$ properties!
In fact, this can be further generalized:
$$\int_0^1 \frac{f(x)}{f(1-x)}\text{d}x=\int_0^1 \frac{f(1-x)}{f(x)}\text{d}x$$
A: As property of definite integrals   it is known that for any function integrand $f(x)$
$$\int_0^a{f(x)}\text{d}x = \int_0^a{f(a-x)}\text{d}x$$
Geometrically this means the area under the curve when flipped about the line $ x= \dfrac{a}{2}$ as a rigid figure cannot change. The property can be stated symbolically as in  one particular generalisation where ${p} $ is a constant:
$$\int_0^1 \frac{f(x)^p}{f(1-x)^p}\text{d}x=\int_0^1 \frac{f(1-x)^p}{f(x)^p}\text{d}x.$$
For p = 1 you can write the integrand  also as a product in another  example.
$$ \int_0^{\pi}{\cos^5 \phi \sin \phi }\, d \phi = \int_0^{\pi}{\sin ^5 \phi \cos \phi }\, d \phi $$
I am sure the generalisation  takes a fanciful attention.Earlier days I would imagine a "curved roof trapezoid " and flip it without change of area under the roof. 
A: So you have an integral:
$$I = \int_0^1 \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x$$
Let's split the area of integration by half:
$$I =
\int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x
+
\int_{\tfrac 12}^1 \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x
$$
and substitute $y=1-x$ in the second term:
$$I =
\int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x
+
\int_{\tfrac 12}^0 \left(\frac{(1-y)^{1-y}}{y^y}-\frac{y^y}{(1-y)^{1-y}}\right)(-\text{d}y)
$$
$$
=\int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x
+
\int_{\tfrac 12}^0 \left(\frac{y^y}{(1-y)^{1-y}}-\frac{(1-y)^{1-y}}{y^y}\right)\text{d}y
$$
Now we swap the limits of integration for $y$:
$$I =
\int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x
-
\int_0^{\tfrac 12} \left(\frac{y^y}{(1-y)^{1-y}}-\frac{(1-y)^{1-y}}{y^y}\right)\text{d}y
$$
and after a few minutes of staring up we see the expression is
$$I = K - K$$
with
$$K = \int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x$$
Hence $I = 0$.
