Cancelling minus sign while integrating I was at this step of integration:
$$ \int p^{-2} \mathrm d p = \int -2x \mathrm dx \,.$$
I then wrote the next step as 
$\frac{-1}{p} = -x^2  +  c$            (my answer).
However the solution for the problem in the differential equation worksheet read
$\frac{1}{p} = x^2  +  c$ (solution key).   
My question is, am I supposed to cancel the minus sign in both LHS & RHS term and then apply the constant c?
 A: Recall that the constant of integration appears in every antiderivative.  But this "$+C$" does not represent a constant, it represents every possible constant, simultaneously, which we might represent by the phrase "any real number".  It is a placeholder for "you can vary this function by any constant you like and its derivative will be the integrand you started with".  So, let's be a little more wordy:  \begin{align}
   \int p^{-2} \,\mathrm{d}p &= \int -2 x \,\mathrm{d}x  \\
   \frac{-1}{p} + \text{ any real number } &= -x^2 + \text{ any real number }  \\
   \frac{1}{p} - \text{ any real number } &= x^2 - \text{ any real number }  \\
   \frac{1}{p} &= x^2 - \text{ any real number } + \text{ any real number }  \\
\frac{1}{p} &= x^2 + \text{ any real number }  \\
\frac{1}{p} &= x^2 + C  \text{.}
\end{align}
Remember:  


*

*minus "any real number" is "any real number"  ($-C = C$), 

*a nonzero constant times "any real number" is "any real number" ($k C = C$), 

*"any real number" plus or minus "any real number" is "any real number" ($C +C = C - C = C$), and 

*"any real number" times "any real number" is "any real number" ($C \cdot C = C$).  


The arithmetic of "$+C$"s is very simple.  As long as you don't try to imagine that "$C$" represents a particular real number you will be in good shape.  (Note that sometimes, due to additional information, we are able to select one of the antiderivatives (i.e., constrain $C$ to a particular real number) that satisfies the additional information.)
A: Both answers are the same in terms of showing the knowledge required to solve the problem, but a teacher might expect you to do your best to simplify your answer. It's more about your teacher's tolerance to un-neatness than mathematical consistency.
