Does $\int_{V_i}^{V_f} P dV = CT \int_{V_i}^{V_f} \frac{1}{V} dV$ imply $P= \frac{CT}{V}$? 
*

*Does equation $(1)$ imply equation $(2)$?

$$\int_{V_i}^{V_f} P dV = CT \int_{V_i}^{V_f} \frac{1}{V} dV \tag{1}$$
$$P= \frac{CT}{V}\tag{2}$$
My motivation for doing so is the following statement from thermodynamics:
$\begin{eqnarray}
\Delta W = P dV &=& cT ln\big(\frac{V_f}{V_i}\big) \rightarrow
P= \frac{\partial W}{\partial V} = \frac{cT}{V}
\end{eqnarray}$


*If I have some expression, for example, $axe^{bx}$ how do I know what function to integration to obtain that? I.e. how can I solve the following for $f(x)$?

$$axe^{bx} = \int f(x) dx$$
Edit: wow.... I am so tempted to remove this embarrassing lapse...
 A: *

*The answer is no. Try a constant $P$. What's the integral of $\frac{dv}v$?
With the new edits, move $CT$ inside the integral on the right, move the entire integral to the other side, and combine the two integrals. You get $$\int_{v_1}^{v_2}\left(P-\frac{CT}v\right)dv=0$$ If this equation is true for all possible integration limits, than equation $2$ is the only solution, assuming $P(v)$ is continuous.


*Use the fundamental theorem of calculus. Take the derivative of both sides
A: Hint for second part of your Q:
$$\left(\int f(x)dx \right)'=f(x).$$
A: *

*Let $I$ be an interval$$\int_I f(x) \, \mathrm{d}x = \int_I g(x) \, \mathrm{d}x$$
does not imply $f(x) = g(x)$ in $I$. Why is that? Hint: Consider, for example $I = [0, 1]$. Can you come up with examples of functions that integrate to $0$?


*Maybe try Googling "antiderivative"?
A: Rather than equation (1) implying equation (2) it is rather the other way around. In this context we want to find the work done (e.g. under the compression of a piston filled with a fluid). As you rightly pointed out:
$$dW=PdV$$
which leads us to:
$$W=\int PdV$$
however without making assumptions about $P$ this is very difficult to do, so we use the equation:
$$P=\frac{CT}{V}$$
which you may recognise as similar to: $pV=nR_0T$ or $pV=mRT$. I think in this context you have confused yourself between volume, $V$ and specific volume $v=\frac 1\rho$ and this is where you can see the assumption that has been made: a fluid of constant volume. This formula is used as:
$$\frac p\rho=pv=RT$$ In this case you have used $C$ rather than $R$ but it is the same thing. Now plugging this back in we get:
$$W=\int Pdv=\int\frac{CT}{v}dv=CT\int \frac {dv}v$$
Hope this helps!
