Let f(x) be a polynomial of degree three satisfying f(0)=-1 and...

Let $f(x)=a x^{3}+b x^{2}+c x+d$ Put $x=0 and x=1$ Then, we get f(0)=-1 and $f(1)=0\Rightarrow \mathrm{d}=-1$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=1$ It is given that $x=0$ is a stationary point of $\mathrm{f}(\mathrm{x})$, but it is not a point of extremum. Therefore, $f(0)=0=f^{\prime \prime}(0) and f^{\prime \prime \prime}(0)=0$ Now, $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}\Rightarrow f(x)=3 a x^{2}+2 b x+c$$\begin{array}{l} f^{\prime \prime}(x)=6 a x+2 b \text{ and } f^{\prime \prime}(x)=6 a \\ f^{\prime}=0, f^{\prime \prime}(0)=0\text{ and } f^{\prime \prime \prime}(0)=0 ≠ 0 \end{array}$$\Rightarrow c=0, b=0 and a \neq 0 From$Eqs. (i) and (ii), we get \mathrm{a}=1$,$\mathrm{ b}=\mathrm{c}=0$and$\mathrm{d}=$-1Put these values in$\mathrm{f}$($\mathrm{x})$we get$f(x)=x^{3}-1$Hence,$\int \frac{\mathrm{f}$($\mathrm{x})}{\mathrmx^{3}-1}\mathrm{dx}=\int \frac{\mathrmx^{3}-1}{\mathrmx^{3}-1} \mathrm{ d} \mathrm{x}=\int 1 \mathrm{ d} $\mathrm{x}=\mathrm{x}$+\mathrm{C}$