Consider:
$$ \int \frac{ \ln (t-1)}{t} dt$$
And then,
$$ I( \alpha) = \int \frac{ \ln [\alpha (t-1)] }{t} dt$$
D.w.r.t. $\alpha$
$$ \frac{dI}{d \alpha} = \int \frac{ 1}{ \alpha t} dt$$
$$ \frac{dI}{d \alpha} = \frac{1}{\alpha} \ln t +C \tag{1}$$
Integrating both side with $\alpha$,
$$ I = \ln \alpha \ln t + C \alpha$$
And putting $ \alpha =1$
I get a weird result...:
$$ I = C$$
What did I do wrong..? Also is the constant in (1) a function of $ \alpha$ ..? (generally speaking)
Ok, the differentiation is where I messed up.
$$ I_{\alpha} = \int \frac{\partial}{\partial \alpha} \frac{ \ln (\alpha (t-1) )}{t} dt$$
Or,
$$I_{\alpha} = \int \frac{1}{t} dt$$
Hence,
$$ I = \ln t \alpha +C$$
Hence,
$$ \int \frac{ \ln [\alpha (t-1)] }{t} dt = \alpha \ln t + C$$
Too bad there is no easy points to evaluate this at >:(