Let $f(x)$ be a polynomial of degree three satisfying $f(0)=-1$ and $f(1)=0$. Also, 0 is a stationary point of $f(x)$. If $f(x)$ does not have an extremum at $x=0$, then the value of the integral $\int \frac{f(x)}{x^3-1} d x$, is

$x+C$

Let $f(x)=a x^3+b x^2+c x+d$. Then,

$f(0)=-1$ and $f(1)=0$

$\Rightarrow d=-1$ and $a+b+c+d=0$

$\Rightarrow d=-1$ and $a+b+c=1$   ... (i)

It is given that x = 0 is a stationary point of f(x) but it is not a point of extremum. Therefore,

$f'(0)=0, f''(0)=0$ and $f'''(0) \neq 0$

Now,

$f(x)=a x^3+b x^2+c x+d$

$\Rightarrow f'(x)=3 a x^2+2 b x+c, f''(x)=6 a x+b$ and $f'''(x)=6 a$

∴  $f'(0)=0, f''(0)=0$ and $f'''(0) \neq 0$

$\Rightarrow c=0, b=0$ and $a \neq 0$         ....(ii)

From (i) and (ii), we get

$a=1, b=c=0$ and $d=-1$

∴   $f(x)=x^3-1$

Hence, $\int \frac{f(x)}{x^3-1} d x=\int 1 . d x=x+C$