$\int_1^e \frac{1 + \log x}{x} \, dx = $

Let $\alpha$ and $\beta$ $(\alpha < \beta)$ be the roots of $18x^2 - 9\pi x + \pi^2 = 0$,$f(x) = x^2$,and $g(x) = \cos x$. Then $\int_{\alpha}^{\beta} x (g \circ f(x)) dx =$

$\int_0^{\pi /4} \tan^6 x \sec^2 x \, dx = $

Let $\int_\alpha^{\log _e 4} \frac{dx}{\sqrt{e^{x}-1}}=\frac{\pi}{6}$. Then $e^\alpha$ and $e^{-\alpha}$ are the roots of the equation :

$\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6} \Rightarrow x=$

$\int_0^1 \frac{1}{2+\sqrt{x}} \, dx =$