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 from each of the three boxes containing $$3$$ white and $$1$$ black, $$2$$ white and $$2$$ black, $$1$$ white and $$3$$ black balls, one ball is drawn at random, then the probability that $$2$$ white and $$1$$ black ball will be drawn is

If $$\overline E $$ and $$\overline F $$ are the complementary events of events $$E$$ and $$F$$ respectively and if $$0 < P\left( F \right) < 1,$$ then