Integration of Force to obtain Potential Energy Given that for a force $F$,
$F=- \frac{GMm}{r^2} + \frac{Am c^2 r}{3}$, 
where A and M are constant, and the potential energy $V$, where 
$V= - \int Fdr$, 
would I be correct in saying that 
$V=- \frac{A c^2 m r^2}{6} + \frac{GMm}{r}$. 
I did try the physics SE for this, but no joy. Thanks!
 A: 
Given:
$$\text{F}=\frac{\text{A}\text{m}\text{c}^2\text{r}}{3}-\frac{\text{G}\text{M}\text{m}}{\text{r}^2}$$


So:
$$\text{V}=-\int\text{F}\space\text{d}\text{r}=-\int\left[\frac{\text{A}\text{m}\text{c}^2\text{r}}{3}-\frac{\text{G}\text{M}\text{m}}{\text{r}^2}\right]\space\text{d}\text{r}=$$
$$\int\frac{\text{G}\text{M}\text{m}}{\text{r}^2}\space\text{d}\text{r}-\int\frac{\text{A}\text{m}\text{c}^2\text{r}}{3}\space\text{d}\text{r}=$$
$$\text{G}\text{M}\text{m}\int\frac{1}{\text{r}^2}\space\text{d}\text{r}-\frac{\text{A}\text{m}\text{c}^2}{3}\int\text{r}\space\text{d}\text{r}=$$
$$\text{G}\text{M}\text{m}\cdot-\frac{1}{\text{r}}-\frac{\text{A}\text{m}\text{c}^2}{3}\cdot\frac{\text{r}^2}{2}+\text{K}=$$
$$-\frac{\text{G}\text{M}\text{m}}{\text{r}}-\frac{\text{A}\text{m}\text{c}^2\text{r}^2}{6}+\text{K}$$
So, we get that:
$$\text{V}=-\int\text{F}\space\text{d}\text{r}=-\frac{\text{G}\text{M}\text{m}}{\text{r}}-\frac{\text{A}\text{m}\text{c}^2\text{r}^2}{6}+\text{K}$$
Where $\text{K}$ is the integration constant.
