$$I = \int {\cos (\ln x)dx} $$. Then I =

Let $$\int {{{{x^{{1 \over 2}}}} \over {\sqrt {1 - {x^3}} }}dx = {2 \over 3}g(f(x)) + c} $$ ; then

(c denotes constant of integration)

If $$\int {{{\sin 2x} \over {{{(a + b\cos x)}^2}}}dx} = \alpha \left[ {{{\log }_e}\left| {a + b\cos x} \right| + {a \over {a + b\cos x}}} \right] + c$$, then $$\alpha$$ is equal to

$$\int {{{f(x)\phi '(x) + \phi (x)f'(x)} \over {(f(x)\phi (x) + 1)\sqrt {f(x)\phi (x) - 1} }}dx = } $$