Value of $\int_{0}^{1}\dfrac{\log x}{1-x}dx$. What is my wrong step? I would like to evaluate the value of this integral:
$$I=\int_{0}^{1}\dfrac{\log x}{1-x}dx.$$

On one hand, I proceed using integration by parts as follows:
$$I=\int_{0}^{1}f(x)g'(x)dx,$$
where $f(x)=\log x$ and $g'(x)=\dfrac{1}{1-x}$. From this,  I can write: $f'(x)=\dfrac{1}{x}$ and $g(x)=-\log(1-x)$.
Therefore:
$$\begin{equation}\begin{split}I=\int_{0}^{1}f(x)g'(x)dx&=\left[f(x)g(x)\right]_{0}^{1}-\int_{0}^{1}f'(x)g(x)dx,\\&=\left[-\log(1-x)\log x\right]_{0}^{1}+\int_{0}^{1}\dfrac{\log(1-x)}{x}dx,\\&=\int_{0}^{1}\dfrac{\log(1-x)}{x}dx\\&=-\int_{0}^{1}\dfrac{\log(t)}{1-t}dt\\&=-I.\end{split}\end{equation}$$
Hence, $I=0.$
On the other hand, Wolframalpha gives $I=-\dfrac{\pi^2}{6}.$

Don't use integration by parts straight away. Instead, expand the denominator: $\sum_{k=0}^{\infty} x^k$ because the bounds on $x$ are strictly between $0$ and $1$. After this, interchange integration and summation (due to uniform convergence and you'll get integrals of the form $\int_{0}^{1} x^k \log x dx$. Now use integration by parts. 
