$1/x$ simple integral with different starting conditions

Question

I simply want to test:

$\displaystyle{\int} \dfrac{1}{T_0\,-\Gamma(h-h_0)}\,\mathrm{dh}$ with $T(h_0) = T_0$

is the same as

$\displaystyle{\int_{h_0}^{h}} \dfrac{1}{T_0\,-\Gamma(h-h_0)}\,\mathrm{dh}$

Actually it should be, but

$\displaystyle{\int} \dfrac{1}{T_0\,-\Gamma(h-h_0)}\,\mathrm{dh} = -\dfrac{\ln(T_0-\Gamma(h-h_0))}{\Gamma} + C \quad \Rightarrow \quad T_0 = -\dfrac{\ln(T_0)}{\Gamma}+C \quad \Rightarrow \quad C = T_0 - \dfrac{\ln(T_0)}{\Gamma}$

whereas

$\displaystyle{\int_{h_0}^{h}} \dfrac{1}{T_0\,-\Gamma(h-h_0)}\,\mathrm{dh} = {\left[-\dfrac{\ln(T_0-\Gamma(h-h_0))}{\Gamma}\right]}^h_{h_0} = -\dfrac{\ln(T_0-\Gamma(h-h_0))}{\Gamma}+\dfrac{\ln(T_0)}{\Gamma}$

or is this natural?