Answer
$$
f(x) =x^{5}-x^{4}+x^{3}+C x+D
$$
where $C,D$ are arbitrary constants .
Work Step by Step
$$
f^{\prime \prime}(x)=20 x^{3}-12 x^{2}+6 x
$$
The general anti-derivative of $
f^{\prime \prime}(x)=20 x^{3}-12 x^{2}+6 x
$ is
$$
\begin{split}
f^{\prime}(x)& =20\left(\frac{x^{4}}{4}\right)-12\left(\frac{x^{3}}{3}\right)+6\left(\frac{x^{2}}{2}\right)+C \\
& =5 x^{4}-4 x^{3}+3 x^{2}+C
\end{split}
$$
Using the anti-differentiation rules once more, we find that
$$
\begin{split}
f(x) &=5\left(\frac{x^{5}}{5}\right)-4\left(\frac{x^{4}}{4}\right)+3\left(\frac{x^{3}}{3}\right)+C x+D \\
&=x^{5}-x^{4}+x^{3}+C x+D
\end{split}
$$
where $C,D$ are arbitrary constants .
