How to integrate $\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx$? 
How to integrate  $\displaystyle\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx$ ? 
If $\theta\in(0,\frac{1}{2})$ 

$$
\begin{align}
& \int_0^{\theta}\frac{x}{(1-x^2)(1-x)}\,dx \\[8pt]
= & \int_0^{\theta}\frac{x}{(1-x)(1+x)(1-x)}\,dx \\[8pt]
= & \int_0^{\theta}\frac{1}{(1-x)^2}\frac{x}{1+x}dx \\[8pt]
= & \int_0^{\theta}\frac{1}{(1-x)^2}\Big(1-\frac{1}{1+x}\Big) \, dx \\[8pt]
= & \int_0^{\theta}\frac{1}{(1-x)^2}-\displaystyle\int_0^{\theta}\frac{1}{(1-x^2)(1-x)} \, dx
\end{align}
$$
LHS is OK, but RHS ?

How about partial fraction decomposition?
$$\begin{align} \displaystyle\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx & = \int_0^\theta \frac x{(1-x)(1+x)(1 - x)} \,dx \\ \\ & = \int_0^\theta \frac x{(1+x)(1-x)^2}dx \\ \\ & = \int_0^\theta \left[ \frac A{1+x} + \frac B{1-x} + \frac C{(1-x)^2}\right]dx\end{align}$$
