Why does one integral converge and not the other? Maybe I am just not thinking this through properly but why does. Or maybe I'm just trusting wolfram alpha too much
(1) Doesn't Converge
$$\int_{0}^{\frac{1}{1.95}} \frac{1}{1-1.95x}dx$$
(2) Does Converge 
$$\int_{0}^{\frac{1}{1.95}} \frac{1.01}{1-1.95x}dx$$
(3) Does Converge 
$$\int_{0}^{\frac{1}{1.95}} \frac{0.99}{1-1.95x}dx$$
I would intuitively think that since (1) doesnt converge and (2) and (3) shouldn't, I may just thinking of this too simply
 A: Since all three integrands are scalar multiples one of another, the three integrals should behave analogously. In particular, they all diverge, since
$$\int_0^{\frac1{1.95}}\frac1{1-1.95x}dx=\lim_{a\to{\frac1{1.95}}^-}\frac1{-1.95}\log(1-1.95x)|_0^a=+\infty,$$
and any constant different from one could be taken out of the integral and wouldn't change the result (except for the zero, and for the sign of $\infty$ if it were negative.)
A: The improper integral is not difficult:
$$
\int_0^{\frac{1}{a}} \frac{y}{1-a x}\, dx
=\lim_{b\to\frac{1}{a}-} \int_0^b\frac{y}{1-a x}\, dx
= \lim_{b\to \frac{1}{a}-}-\frac{y}{a}\ln(1-a x)\bigg|_{x=0}^{b}\tag{1}
$$
where $a=1.95$.
Unless $y=0$, the limit in (1) does not converge.
A: This definitely seems to be a bug in Wolfram Alpha.  If you look at computation lower on the page, it gives the Cauchy principal value of the integral as $16.5433$  Apparently, when the numerator is not $1$, it's reporting the Cauchy principal value as the value of the (divergent) integral.
