Finding expectation of log Does somebody have an idea to find the following expectation:
$$E(X\log(a+bX)),$$
How can I proceed, can we go like this:
$$=E(X)E[\log(a+bX)],$$if yes then what?
 A: With appropriate restrictions on $a,b$, you may Taylor expand $\log(a+b\,x)$ around $x=0$ to get
$$
  \log(a+b\,x) = \log a + \frac{b}{a}\,x - \frac{b^2}{2a^2} \,\,x^2 + O(x^3) \, .
$$
The series converges for $|x|<|a/b|$, giving the approximation
$$
  \mathrm{E}[X \log(a +b\,X)] \approx (\log a) \, \mathrm{E}[X] + \frac{b}{a} \mathrm{E}[X^2] \, .
$$
Finally, remember that $\mathrm{E}[X^2]=\mathrm{Var}[X] + \mathrm{E}^2[X]$.
A: The expectation is $\int_\Omega x\log(a+bx)dF$ where F is the cumulative distribution function, $\Omega$ is the sample space and $\int$ is the Lebesgue integral. (In the real world, this typically means it's either a sum or an integral, depending on whether your distribution is discrete or continuous.)
If you are starting from a sample, you would simply estimate the expectation in the usual way: $$\frac{1}{n}\sum_{i=1}^n X_i\log(a+bX_i)$$
If you know the distribution, you can attempt to integrate it. It looks messy, so you will probably need to use numeric methods. 
