Easy way to compute the area between $f(x)=x$ and $g(x)=x^2\ln(x)$ Is there an easy to compute the area between $f(x)=x$ and $g(x)=x^2\ln(x)$ without refering to the Lambert W-function?
 A: By letting $x_0$ be the positive point that satisfies $x_0 \log x_0 = 1$, we get that the relevant integral equals a polynomial in $x_0$ (of degree $3$). This is probably the simplest expression you can get for this value. Since $x_0 = W(1)^{-1}$, I would guess the answer to your question is "no".
A: By looking at a figure we can assume that the functions intersects at some point $x_0$, which we will write as $x_0 = \exp(\Omega)$ for reasons that will become apparent later. 
Again from figure we obtain that the enclosed area can be expressed as
$$ \begin{align} 
A & = \int_0^{e^{\Omega}} x - x^2 \log x\,\mathrm{d}x \\ 
  & = \frac{1}{2} e^{2 \Omega} - \frac{1}{3} \Omega e^{3\Omega} + \frac{1}{9} e^{3 \Omega} 
\end{align} $$
Now if we just could find $\Omega$ we would be done. But luckilly $\Omega$ is a "well" know constant!
http://en.wikipedia.org/wiki/Omega_constant
Using some exponent rules $a^b a^b= a^{2b}$, and that $1 = \Omega \exp(\Omega)$,
we can rewrite the area as
$$ A = \frac{1}{6} e^{2W} + \frac{1}{9} e^{3W} $$
For example by using the recurrence relation
$$ \Omega_{n+1} = \frac{1 + \Omega_n}{1 + e^{\Omega_n}} $$
With the initial estimate of $\log 2$. We get that 
$$ \Omega \approx \frac{1 + \log 2}{3}$$
Which is just $0.0027$ from the real value, replacing $\Omega$ with this expression the area can be rewritten as
$$ A = \frac{1}{6} 2^{2/3} e^{2/3} + \frac{2e}{9} \approx 1.1272$$
Which is quite close to the actuall value.
