If $$\int_\limits0^a \sqrt{\frac{a-x}{x}} d x=\frac{k}{2}$$, then $$k=$$

$$\int\limits_{ - \pi }^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}dx = } $$

The value of $$\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{2x - 1} \over {1 + x - {x^2}}}} \right)dx} $$ is

$$\int\limits_5^{10} \frac{d x}{(x-1)(x-2)}=$$