$\int \frac{d x}{\sqrt{x}(x+9)}$ is equal to

If $\int \frac{\sqrt{1-x^2}}{x^4} \,dx = A(x)\left(\sqrt{1-x^2}\right)^{m} + c$ for a suitably chosen integer $m$ and a function $A(x)$,where $c$ is a constant of integration,then $(A(x))^{m}$ equals

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 = $

The integral $\int {\frac{{dx}}{{{{(x + 1)}^{\frac{3}{4}}}{{(x - 2)}^{\frac{5}{4}}}}}} $ is equal to

Integrate the function $\frac{1}{x(\log x)^{m}}$,where $x > 0$ and $m \neq 1$.

$\int {\frac{{\sec x(1 + \tan x)dx}}{{({e^{ - x}} + \sec x)}}} = f(x) + C$ where $f(0) = \ln 2$,then $f\left( {\frac{\pi }{4}} \right)$ is -