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

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

An n-digit number is a positive number with exactly digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is