Evaluating $\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$ At first sight it looks like the integral below
$$\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$$
can be evaluated by using some geometric series. What else can we do? Is there a fast easy way
to finish it?
 A: Sorry for the previous wrong answer. Since Mr. @robjohn has answered this OP correctly and nicely, I'd not try to answer this OP because it'll just look like I copy his answer. Therefore, I'm going to propose another way to evaluate the first RHS integral in $(1)$, refer to Mr. @robjohn's answer, instead of using the limit of a difference of beta functions.
Consider the following integral
\begin{align}
I(s)&=\int_0^1 \frac{1-t^s}{1-t}\,dt\\
I'(s)&=-\int_0^1  \frac{t^s\ln t}{1-t} \,dt\\
&=-\int_0^1\sum_{n=0}^\infty t^{n+s}\ln t\,dt\\
&=-\sum_{n=0}^\infty\partial_s\int_0^1 t^{n+s}\,dt\\
&=-\sum_{n=0}^\infty\partial_s\left[\frac{1}{n+s+1}\right]\\
&=\sum_{n=0}^\infty\frac{1}{(n+s+1)^2}\\
&=\psi_1(s+1)\\
I(s)&=\int\psi_1(s+1)\,ds\\
&=\int\frac{\partial}{\partial s}\bigg[\psi(s+1)\bigg]\,ds\\
&=\psi(s+1)+C\\
\end{align}
For $s=0$, we have $I(0)=0$. Implying $C=-\psi(1)=\gamma$, then
\begin{align}
I(s)=\int_0^1 \frac{1-t^s}{1-t}\,dt=\psi(s+1)+\gamma
\end{align}
Thus
\begin{align}
\int_0^1 \left[\frac{t^{a-1}}{1-t}-\frac{t^{\large\frac{b}{c}-1}}{1-t}\right]\ dt
&=\int_0^1 \left[\frac{1-t^{\large\frac{b}{c}-1}}{1-t}-\frac{1-t^{a-1}}{1-t}\right]\ dt\\
&=\psi\left(\frac{b}{c}\right)-\psi\left(a\right)
\end{align}
It holds for $\Re\left(a\right)>0$, $\Re\left(b\right)>0$, and $\Re\left(c\right)>0$.
A: $$
\begin{align}
\lim_{d\to1^-}\int_0^d\left[\frac{t^{a-1}}{1-t}-\frac{c\,t^{b-1}}{1-t^c}\right]\mathrm{d}t
&=\lim_{d\to1^-}\int_0^d\left[\frac{t^{a-1}}{1-t}-\frac{t^{b/c-1}}{1-t}\right]\mathrm{d}t\tag{1}\\
&-\lim_{d\to1^-}\int_d^{d^c}\frac{t^{b/c-1}}{1-t}\mathrm{d}t\tag{2}\\
&=\lim_{d\to1^-}\int_0^1\left[\frac{t^{a-1}}{(1-t)^d}-\frac{t^{b/c-1}}{(1-t)^d}\right]\mathrm{d}t\tag{3}\\
&-\lim_{d\to1^-}\log\left(\frac{1-d}{1-d^{c}}\right)\tag{4}\\
&=\lim_{d\to1^-}\mathrm{B}(a,1-d)-\mathrm{B}(b/c,1-d)\tag{5}\\
&+\log(c)\tag{6}\\
&=\lim_{d\to0^+}\left[\frac{\Gamma(a)\Gamma(1+d)}{d\,\Gamma(a+d)}-\frac{\Gamma(b/c)\Gamma(1+d)}{d\,\Gamma(b/c+d)}\right]\tag{7}\\
&+\log(c)\tag{8}\\
&=\boxed{\displaystyle\bbox[5px]{\frac{\Gamma'(b/c)}{\Gamma(b/c)}-\frac{\Gamma'(a)}{\Gamma(a)}+\log(c)}}\tag{9}
\end{align}
$$
Explanation:
$(1)$: to substitute on the second part of the difference, we must break up the integral. To do so, we must reduce the domain of integration to $[0,d]$.
$(2)$: The substitution in $(1)$ changes the upper limit of the right term to $[0,d^c]$. We add this term to make up for this change in domain.
$(3)$: the integrand in $(1)$ is now bounded (it tends  to $b/c-a$ as $t\to1$), However, we cannot evaluate each term separately; but, we can if we divide by $(1-t)^d$ and let $d\to1^-$ (using the dominated convergence theorem).
$(4)$: the integral in $(2)$ is between $d^{b/c-1}\to1$ and $d^{b-c}\to1$ times $\int_d^{d^c}\frac{\mathrm{d}t}{1-t}=\log\left(\frac{1-d}{1-d^{c}}\right)$
$(5)$: The integral in $(3)$ is now the difference of two Beta functions
$(6)$: evaluate the limit in $(4)$
$(7)$: substitute $d\mapsto1-d$ in $(5)$ then use $d\,\Gamma(d)=\Gamma(1+d)$
$(8)$: copy $(6)$
$(9)$: subtract and take the limit in $(7)$ and add $(8)$

In this answer, it is shown that
$$
\frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1}\right)\tag{10}
$$
In the spirit of Anastasiya-Romanova's answer, let us evaluate
$$
\begin{align}
\int_0^1\frac{1-t^{x-1}}{1-t}\mathrm{d}t
&=\sum_{k=1}^\infty\int_0^1(t^{k-1}-t^{k+x-2})\,\mathrm{d}t\\
&=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1}\right)\,\mathrm{d}t\\
&=\frac{\Gamma'(x)}{\Gamma(x)}+\gamma\tag{11}
\end{align}
$$
We can now use $(11)$ to evaluate
$$
\begin{align}
\int_0^1\left[\frac{t^{a-1}}{1-t}-\frac{t^{b/c-1}}{1-t}\right]\mathrm{d}t
&=\int_0^1\left[\frac{1-t^{b/c-1}}{1-t}-\frac{1-t^{a-1}}{1-t}\right]\mathrm{d}t\\
&=\frac{\Gamma'(b/c)}{\Gamma(b/c)}-\frac{\Gamma'(a)}{\Gamma(a)}\tag{12}
\end{align}
$$
to which we still need to add $\log(c)$ to get the answer above.
A: The solution in the book, (Almost) Impossible Integrals, Sums, and Series (pages $210$-$211$), presented in large steps :
Using Digamma integral representation, $\displaystyle\psi(x)=\int_0^{\infty} \left(\frac{e^{-t}}{t}-\frac{e^{-x t}}{1-e^{-t}}\right)  \textrm{d}t$, we have 
$$\int_0^1 \left(\frac{x^{a-1}}{1-x}-\frac{c x^{b-1}}{1-x^c}\right)  \textrm{d}x\overset{x= e^{-y}}{=}\int_0^{\infty} \left(\frac{e^{-ay}}{1-e^{-y}}-\frac{c e^{-b y}}{1-e^{-c y}}\right)  \textrm{d}y$$
$$=\int_0^{\infty} \left(\frac{e^{-c y}}{y}-\frac{c e^{-b y}}{1-e^{-c y}}-\left(\frac{e^{-y}}{y}-\frac{e^{-a y}}{1-e^{-y}}\right)+\frac{e^{-y}-e^{-c y}}{y}\right)  \textrm{d}y$$
$$=\underbrace{\int_0^{\infty} \left(\frac{e^{-c y}}{y}-\frac{c e^{-b y}}{1-e^{-c y}}\right)  \textrm{d}y}_{\displaystyle \psi(b/c)}-\underbrace{\int_0^{\infty} \left(\frac{e^{-y}}{y}-\frac{e^{-a y}}{1-e^{-y}}\right) \textrm{d}y}_{\displaystyle \psi(a)}+\underbrace{\int_0^{\infty}\frac{e^{-y}-e^{-c y}}{y} \textrm{d}y}_{\log(c) - \text{Frullani's integral}}$$
$$=\psi\left(\frac{b}{c}\right)-\psi(a)+\log(c).$$
The integral is originally created by Srinivasa Ramanujan as mentioned in the book. 
A supplementary result: If we set in Ramanujan's integral $a=s$, $b=s$, $c=2$, we get that
\begin{equation*}
\begin{aligned}
\psi\left(\frac{s}{2}\right)-\psi(s)+\log(2)&=\int_0^{1} \left(\frac{x^{s-1}}{1-x}-\frac{2 x^{s-1}}{1-x^2}\right) \textrm{d}x\\
                                                                   &=\int_0^{1} \left(\frac{x^{s-1}}{1-x}-x^{s-1}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\right) \textrm{d}x\\
                                                                   &=-\int_0^{1} \frac{x^{s-1}}{1+x}\textrm{d}x,
\end{aligned}
\end{equation*}
whence we obtain

$$\int_0^{1} \frac{x^{s-1}}{1+x}\textrm{d}x=\psi(s)-\psi\left(\frac{s}{2}\right)-\log(2),$$

and this proves differently the second closed-form of the integral in Section $1.5$, the result in $(1.10)$, page $3$, from the mentioned book. Full credit 
 and many thanks go to Cornel.
