Integrating V(x)DV(X) in NMOS Transistors: A Mathematical Analysis

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I'm in an electronics course, and the book derives an equation for the current-voltage characteristics of an NMOS transistor. In doing so, it integrate this:
<br /> \int_{0}^{V_{DS}}V(X)\, dV(X)=\frac{V_{DS}^2}{2}I can see that integrating a function F(X) with respect to F(X) turns out to be the same as integrating a single variable such as x with respect to x, but is that mathematically kosher? Can someone convince me that it is?
 
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To see that the antiderivative is

\int V(x)dV(x) = \frac{V^2(x)} 2+ C

Just note that

\int V(x)dV(x) = \int V(x)V&#039;(x)dx

and that integrand is exactly what you get if you differentiate

\frac{V^2(x)} 2+ C
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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