$\int \frac{dx}{x\sqrt{1 - (\log x)^2}} = $

$\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \,dx = \frac{1}{2}(g(x))^2 + C$,(where $C$ is the constant of integration). Then $g(x) =$

If $l^r(x)$ denotes the $r$-th iterated logarithm of $x$,i.e.,$l^1(x) = \log(x)$,$l^2(x) = \log(\log(x))$,...,$l^r(x) = \log(\log(...\log(x)...))$,then $\int \frac{1}{x \cdot l^1(x) \cdot l^2(x) \cdot ... \cdot l^r(x)} \, dx = $

$\int \frac{d x}{\cos x \sqrt{\cos 2 x}} = $

If $\int \frac{\sqrt[4]{x}}{\sqrt{x}+\sqrt[4]{x}} d x=\frac{2}{3}\left[A \sqrt[4]{x^3}+B \sqrt[4]{x^2}+C \sqrt[4]{x}+D \log (1+\sqrt[4]{x})\right]+K$ then $\frac{2}{3}(A+B+C+D)=$

$\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} d x=$