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In solving the Van't Hoff equation

$$ \frac{d\ln(K)}{dT}=\frac{\Delta H^{o}}{RT^{2}} \,\, . $$

Considering the term $\Delta H^{o}$ constant, I usually perform the separation o variables and then I perfrom an integration over the interval $[T_{1},T_{2}]$

$$ \int^{T_{2}}_{T_{1}}\frac{d\ln(K)}{dT}dT=\frac{\Delta H^{o}}{R}\int^{T_{2}}_{T_{1}}T^{-2}dT \,\, . $$

My doubt lies in which is the formal way to perform a change of variables in the left hand of this equation (I'm avoiding to face that differential term as being a fraction and manipulate this such as a fraction).

My second question is how can I know if the better way to integrate this equation is defining the integration limits or just take a indefinite integral? I mean, finding the antiderivative and then findig what's the C, or defining the integration limits?

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By the fundamental theorem of calculus, for any differentiable function $f$, $$ \int_a^bf'(x)dx = f(b) - f(a). $$ Writing the LHS of your equation in a more suggestive form gives $$ \int_{T_1}^{T_2} (\ln K)'(T) dT = \ln K(T_2) - \ln K(T_1). $$ In a more general sense, a change of variables in integration is justified by this and the chain rule $$ \int_a^b f'(u(x))u'(x)dx = \int_a^b(f\circ u)'(x)dx = f(u(b)) - f(u(a)) = \int_{u(a)}^{u(b)} f'(u)du. $$ (Note the slight abuse of notation by using $u$ as the dummy variable in the last integral.) This is the rigorous justification for "cancellation" of differentials that appears when writing in Leibniz notation.

As for your second question, I'm not sure how you'd do the definite integral without finding an antiderivative for the integrand first. And since "indefinite integration" is just finding an antiderivative, I don't think these approaches are actually different.

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  • $\begingroup$ I'm sorry about my second question @eyeballfrog , it wasn't well explained, I fixed in the OP. $\endgroup$ Jun 27, 2017 at 21:44
  • $\begingroup$ Oh. Well that usually depends on your boundary conditions. If it's of the form $f(x_0) = y$, (which it usually is for first-order ODEs) I would suggest just doing a definite integral starting at $x_0$. Otherwise, leaving it as $C$ and figuring it out later is probably fine. $\endgroup$ Jun 27, 2017 at 21:52

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