This function is not invertible .

Let, x>02. Find a function g such that:

integral sign x, 0 tg(t)dt= x+x^2

f(x)=INTEGRAL (from 0 to x) t*g(t) dt

We are told that,

f(x)=x+x^2

Take the derivative of both sides (apply FT of Calculus).

f'(x)=x*g(x)

f'(x)=1+2x

Thus,

1+2x=x*g(x)

Thus,

1/x+2=g(x)