$\int \frac{dx}{x \log x \log(\log x)} = $

$\int \frac{x+1}{(x-2) \sqrt{1-x}} d x=$

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

To find the value of $\int \frac{dx}{x\sqrt{2ax - x^2}}$,the suitable substitution is

$\int \sec^p x \tan x \, dx = $

$\int \frac{\sec^{8} x}{\text{cosec} x} dx =$