If $$\int_0^x {f\left( t \right)dt = x + \int_x^1 {t\,\,f\left( t \right)\,\,dt,} } $$ then the value of $$f(1)$$ is

Let $$f\left( x \right) = x - \left[ x \right],$$ for every real number $$x$$, where $$\left[ x \right]$$ is the integral part of $$x$$. Then $$\int_{ - 1}^1 {f\left( x \right)\,dx} $$ is

Prove that $$\int_0^1 {{{\tan }^{ - 1}}} \,\left( {{1 \over {1 - x + {x^2}}}} \right)dx = 2\int_0^1 {{{\tan }^{ - 1}}} \,x\,dx.$$
Hence or otherwise, evaluate the integral
$$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $$

The order of the differential equation whose general solution is given by
$$y = \left( {{C_1} + {C_2}} \right)\cos \left( {x + {C_3}} \right) - {C_4}{e^{x + {C_5}}},$$ where
$${C_1},{C_2},{C_3},{C_4},{C_5},$$ are arbitrary constants, is