# 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 $$\displaystyle \int_{1}^{\infty} \left( \dfrac{1}{\left \lfloor x \right \rfloor} - \dfrac{1}{x}\right)\, dx$$
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