A problem with integral constants and a differential equation 
This is the expression I have to solve for the axial velocity, represented by  .
Rearranging the expression I get:

After integration:

But my teacher ignores the brackets and comes to:

This is creating a lot of trouble because with brackets the problem becomes wrong...
I am supposed to get to:

,  and  are constants.
The delta sign is equivalent to the Leibnitz notation used before.
Thank you.
 A: Well, when we know that the relation between $\mathcal{P}\left(\text{z}\right)$ and $\text{V}_\text{z}\left(\text{r}\right)$, looks like:
$$\mathcal{P}'\left(\text{z}\right)=\mu\cdot\left\{\frac{1}{\text{r}}\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}\cdot\text{V}_\text{z}'\left(\text{r}\right)\right)\right\}$$
To simplify it:
$$\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}\cdot\text{V}_\text{z}'\left(\text{r}\right)\right)=\text{V}_\text{z}'\left(\text{r}\right)+\text{r}\cdot\text{V}_\text{z}''\left(\text{r}\right)$$
So, we get:
$$\mathcal{P}'\left(\text{z}\right)=\mu\cdot\left\{\frac{1}{\text{r}}\cdot\left(\text{V}_\text{z}'\left(\text{r}\right)+\text{r}\cdot\text{V}_\text{z}''\left(\text{r}\right)\right)\right\}=\mu\cdot\left(\frac{\text{V}_\text{z}'\left(\text{r}\right)}{\text{r}}+\text{V}_\text{z}''\left(\text{r}\right)\right)$$
Let $\text{y}\left(\text{r}\right)=\text{V}_\text{z}'\left(\text{r}\right)$:
$$\mathcal{P}'\left(\text{z}\right)=\mu\cdot\left(\frac{\text{y}\left(\text{r}\right)}{\text{r}}+\text{y}'\left(\text{r}\right)\right)\space\Longleftrightarrow\space\int\frac{\text{d}}{\text{d}\text{r}}\left(\left|\text{r}\right|\cdot\text{y}\left(\text{r}\right)\right)\space\text{d}\text{r}=\int\frac{\mathcal{P}'\left(\text{z}\right)\cdot\left|\text{r}\right|}{\mu}\space\text{d}\text{r}$$
Which leads to (assuming a real valued function):
$$\left|\text{r}\right|\cdot\text{y}\left(\text{r}\right)=\frac{\mathcal{P}'\left(\text{z}\right)}{\mu}\int\left|\text{r}\right|\space\text{d}\text{r}=\frac{\mathcal{P}'\left(\text{z}\right)}{\mu}\cdot\frac{\text{r}^2\cdot\text{sgn}\left(\text{r}\right)}{2}+\text{K}\space\Longleftrightarrow\space$$
$$\text{y}\left(\text{r}\right)=\frac{1}{\left|\text{r}\right|}\cdot\frac{\mathcal{P}'\left(\text{z}\right)}{\mu}\cdot\frac{\text{r}^2\cdot\text{sgn}\left(\text{r}\right)}{2}+\text{K}$$
So, for $\text{V}_\text{z}\left(\text{r}\right)$ we get:
$$\text{V}_\text{z}\left(\text{r}\right)=\int\left\{\frac{1}{\left|\text{r}\right|}\cdot\frac{\mathcal{P}'\left(\text{z}\right)}{\mu}\cdot\frac{\text{r}^2\cdot\text{sgn}\left(\text{r}\right)}{2}+\text{K}\right\}\space\text{d}\text{r}=\frac{\mathcal{P}'\left(\text{z}\right)}{2\mu}\cdot\int\frac{\text{r}^2\cdot\text{sgn}\left(\text{r}\right)}{\left|\text{r}\right|}\space\text{d}\text{r}+\text{K}\cdot\text{r}$$ 
