Integral calculation: residue theorem or it is useless?

Question

I would like to calculate the following integral: $$\int_{0}^{1}dx\frac{x(1-x)^2}{(1-x)^2+ax},$$ where $a$ is a parameter which defined as the integral converges. I would like to try use residue theorem. Thus, I start from the analysis of integrand poles: $$(1-x)^2+ax=0\rightarrow x_{1,2}=1-\frac{a}{2}\pm\frac{1}{2}\sqrt{a^2-4a}.$$ But then I do not understand which contour I should use to evaluate the integral....

Answer 1

Locating the poles is (very) useful, but the residue theorem not so much.
Let us assume that $\nu,\bar{\nu}$ are such that $$ (1-x)^2+ax = (1-\nu x)(1-\bar{\nu}x), $$ i.e. solutions of $\nu+\bar{\nu}=2-a, \nu\bar{\nu}=1$. Then $$ \frac{x(1-x)^2}{(1-x)^2+ax}=x-\frac{ax^2}{(1-x)^2+a}=(x-a)+\frac{a-(2a-a^2)x}{(1-\nu x)(1-\bar{\nu}x)} $$ decomposes as $$ (x-a)+\frac{1}{\bar{\nu}-\nu}\left[\frac{2a-a^2-a\nu}{1-\nu x}-\frac{2a-a^2-a\bar{\nu}}{1-\bar{\nu}x}\right] $$ whose integral over $(0,1)$ equals $$ \frac{1}{2}-a+\frac{a}{\bar{\nu}-\nu}\left[(\nu+a-2)\bar{\nu}\log(1-\nu)-(\bar{\nu}+a-2)\nu\log(1-\bar{\nu})\right]$$ as soon as $\nu,\bar{\nu}\not\in\mathbb{R}$ and $\text{Re}(\nu),\text{Re}(\bar{\nu})<1$.