<s>[INST] Integral $\int^{1}_{1/2}\frac{\ln(x)}{1-x}dx$ 
Evaluate $$\int^{1}_{1/2}\frac{\ln(x)}{1-x}dx$$

Try: Let $$ I =\int^{1}_{1/2}\frac{\ln x}{1-x}dx=\int^{1}_{1/2}\sum^{\infty}_{k=0}x^k\ln(x)dx$$
$$I =\sum^{\infty}_{k=0}\int^{1}_{1/2}x^k\ln(x)dx$$
Using integration by parts, gives: $$I =\sum^{\infty}_{k=0}\frac{\ln(x)x^{k+1}}{k+1}\bigg|^{1}_{1/2}-\sum^{\infty}_{k=0}\frac{x^{k+1}}{(k+1)^2}\bigg|^{1}_{1/2}$$
Can someone help me to solve it? 
 [/INST] Here we will address the integral:
\begin{equation}
I = \int_{\frac{1}{2}}^1 \frac{\ln(x)}{1 - x}\:dx
\end{equation}
We first observe that:
\begin{equation}
\frac{\partial}{\partial a} x^a = x^a \ln(x) \Longrightarrow \ln(x) = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a 
\end{equation}
Thus, 
\begin{equation}
I = \int_{\frac{1}{2}}^1 \frac{\ln(x)}{1 - x}\:dx = \int_{\frac{1}{2}}^1 \frac{\lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a }{1 - x}\:dx
\end{equation}
Which by Leibniz's Integral Rule becomes:
\begin{equation}
I =  \int_{\frac{1}{2}}^1 \frac{\lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a }{1 - x}\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a}  \int_{\frac{1}{2}}^1 \frac{x^a}{1 - x}\:dx
\end{equation}
We now employ the Taylor Series for $\frac{1}{1 - x}$ (which is convergent on the bounds of the integral):
\begin{equation}
\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^n
\end{equation}
Thus our integral becomes:
\begin{align}
I &=  \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a}  \int_{\frac{1}{2}}^1 \frac{x^a}{1 - x}\:dx =  \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \int_{\frac{1}{2}}^1 x^a\sum_{n = 0}^{\infty} x^n \:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \int_{\frac{1}{2}}^1 x^{a + n} \:dx \nonumber \\
&= \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \left[ \frac{x^{a + n + 1}}{a + n + 1}\right]_{\frac{1}{2}}^1 = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \left[ \frac{1}{a + n + 1} - \frac{\left(\frac{1}{2}\right)^{a + n + 1}}{a + n + 1}\right] \nonumber \\
&= \lim_{a \rightarrow 0^+} \sum_{n = 0}^{\infty} \left[ \frac{-1}{\left(a + n + 1\right)^2} - \frac{\left(\frac{1}{2}\right)^{a + n + 1}}{\left(a + n + 1\right)^2} + \frac{\ln\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{a + n + 1}}{a + n + 1}\right] \nonumber \\
&= \sum_{n = 0}^{\infty} \left[\frac{-1}{\left(n + 1\right)^2} - \frac{\left(\frac{1}{2}\right)^{n + 1}}{\left( n + 1\right)^2} + \frac{\ln\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{n + 1}}{n + 1}\right] \nonumber \\
&= -\sum_{n = 0}^{\infty} \frac{1}{(n + 1)^2} - \frac{1}{2}\sum_{n = 0}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{(n + 1)^2} + \frac{\ln\left(\frac{1}{2}\right)}{2}\sum_{n = 0}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n + 1} \nonumber \\
&= -\Phi\left(1,1,2 \right) - \frac{1}{2}\Phi\left(\frac{1}{2},1,2 \right) + \ln(\sqrt{2})\Phi\left(\frac{1}{2},1,1 \right)
\end{align}
Where $\Phi(\cdot, \cdot, \cdot, \cdot)$ is the Lerch Transcedent Function. 
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