Answer
$$
f(x) =Cx+\frac{9}{4}x^{\frac{4}{3}}+\frac{9}{40}x^{\frac{8}{3}}+D
$$
where $C,D$ are arbitrary constants .
Work Step by Step
$$
f^{\prime \prime}(x)= x^{\frac{2}{3}}+ x^{-\frac{2}{3}}
$$
The general anti-derivative of $
f^{\prime \prime}(x)= x^{\frac{2}{3}}+ x^{-\frac{2}{3}}
$ is
$$
f^{\prime}(x)=\frac{3}{5}x^{\frac{5}{3}}+3x^{\frac{1}{3}}+C
$$
Using the anti-differentiation rules once more, we find that
$$
f(x) =Cx+\frac{9}{4}x^{\frac{4}{3}}+\frac{9}{40}x^{\frac{8}{3}}+D
$$
where $C,D$ are arbitrary constants .
