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) does not have an extremum at x = 0, then the value of the integral  $\int \frac{f\left(x\right)}{x^3-1}dx\text{ is }a{x}^b+c$

Let $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$

$\begin{array}{l}\mathrm{f}\left(0\right)=-1\Rightarrow \mathrm{d}=-1\\ \mathrm{f}\left(1\right)=0\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}-1=0\\ \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=1\end{array}$

O is stationary point $\Rightarrow {\mathrm{f}}^{\text{'}}\left(0\right)=0$

$\begin{array}{l}\Rightarrow 3\mathrm{a}(0{)}^2+2\mathrm{b}(0)+\mathrm{c}(1)=0\\ \mathrm{c}=0\end{array}$

f(x) doesnot have an extremum at x = 0

$\begin{array}{l}\Rightarrow {\mathrm{f}}^{\prime \prime }\left(0\right)=0\\ \Rightarrow 6\mathrm{ax}+2\mathrm{b}\left(1\right)=6\mathrm{a}\left(0\right)+2\mathrm{b}=0\\ \mathrm{b}=0\\ \therefore \mathrm{a}=1\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-1\\ \int \frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^3-1}\mathrm{dx}=\int \frac{{\mathrm{x}}^3-1}{{\mathrm{x}}^3-1}\mathrm{dx}=\int 1\mathrm{dx}.\end{array}$

                                          $\begin{array}{l}=\mathrm{x}+\mathrm{c}\\ ={\mathrm{ax}}^{\mathrm{b}}+\mathrm{c}\end{array}$

$\Rightarrow \mathrm{a}=1,\mathrm{b}=1$