The function $L(x) = \int_1^x \frac{dt}{t}$ satisfies the equation

Let the domain of the function $f(x) = \log_2 \log_4 \log_6(3 + 4x - x^2)$ be $(a, b)$. If $\int_0^{b-a} [x^2] dx = p - \sqrt{q} - \sqrt{r}$,where $p, q, r \in \mathbb{N}$,$\gcd(p, q, r) = 1$,and $[\cdot]$ is the greatest integer function,then $p + q + r$ is equal to

$\int_a^b \frac{\log x}{x} \, dx = $

$\int_0^a x^2 (a^2 - x^2)^{3/2} dx = $

Let $\alpha \in (0,1)$ and $\beta = \log_{e}(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^n}{n}$ for $x \in (0,1)$. Then the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to

$\int_0^{\pi /4} \frac{dx}{\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x} = $