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If x ( x -1x 1-x2+ -1x(1-x2)3/2+ x1-x2 )dx=g(x)+C, where C is the constant of integration, then g ( 12 ) equals :
(1) 6 e3
x[f+f']dx=exf with f= x -1x 1-x2, since f'= -1x(1-x2)3/2+ x1-x2 (the other two terms)
g(x)= x ex -1x 1-x2
g ( 12 )= 12 e1/2 6 32= 6 e3
Answer: (1)
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