# Too many parts, part 2 [pun intended] (My fifteenth integral problem)

**Calculus**Level 5

\[ \large \displaystyle \int_{0}^{1} x^{2} 2^{x} \ln x \, dx \]

If the above integral can be expressed in the highly unusual form

\[ \dfrac{-a \, \text{li}(b) + c +d \gamma + \ln \left( \dfrac{\ln^{f} (g)}{h} \right)}{\ln^{j}(k)} \]

where \(a,b,c,d,f,g,h,j,k \) are positive integers and the above expression is in simplest form, find \( a+b+c+d+f+g+h+j+k \).

**Notations**:

- \( \text{li}(x) \) is the logarithmic integral function. \( \text{li}(x) = \displaystyle \int_{0}^{x} \dfrac{dt}{\ln(t)} \)
- \( \gamma \) is the Euler-Mascheroni constant, which may be represented as either \( \displaystyle \lim_{n \to \infty} \left( -\ln(n) + \sum_{k=1}^{n} \dfrac{1}{k} \right) \) or \( \displaystyle \int_{1}^{\infty} \left( \dfrac{1}{\left \lfloor x \right \rfloor} - \dfrac{1}{x}\right)\, dx \)