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Let g(x) be the inverse of the function f(x), and f'(x)= 1/1+x3. then g(x) is equal to....
f(x) = ∫1/(1+x3)dx 1/(1+x3) = (x+1)/3 - (x/3)/(((x-0.5)^2) +0.75) +(2/3)/(((x-0.5)^2) +0.75)
= (x+1)/3 - (1/6)(2x-1)/(((x-0.5)^2) +0.75) - (1/2)(((x-0.5)^2) +0.75)
therefore, ∫1/(1+x3)dx = f(x) = (1/3)ln(x+1) - (1/6)ln(x2-x+1) -(1/√3)arctan((2x-1)/√3)
g(x) = ƒ-1(x) implies that f(g(x))= x
x = (1/3)ln(g(x)+1) - (1/6)ln((g(x))2-x+1) -(1/√3)arctan((2g(x)-1)/√3)
One has to solve this equation in g(x) in order to express g(x) explicitly. It is obvious that this equation cannot be solved.
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