## Calculus 8th Edition

$$f(x) =Cx+\frac{9}{4}x^{\frac{4}{3}}+\frac{9}{40}x^{\frac{8}{3}}+D$$ where $C,D$ are arbitrary constants .
$$f^{\prime \prime}(x)= x^{\frac{2}{3}}+ x^{-\frac{2}{3}}$$ 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 .