evaluate $\int_{0}^{1}dx \frac{(1-x)x(1+x)}{x^2+(1-x)a^2}$

Although probably overkill if you just want the approximate behavior for $a \approx 0$, you can obtain an exact solution for the integral. Notice that $$ \frac{(1-x)x(1+x)}{x^2+(1-x)a^2} = -(x+a^{2})-\left(\frac{a^{4}-a^{2}-1}{2}\right)\frac{2x-a^{2}}{x^{2}-a^{2}x+a^{2}}-\left(\frac{a^{6}-3a^{4}-a^{2}}{2}\right)\frac{1}{\left(x-\frac{a^{2}}{2}\right)^{2}+\left(a\sqrt{1-\frac{a^{2}}{4}}\right)^{2}} $$ which leads you to your integral being equal to $$ \boxed{\int_{0}^{1} \frac{(1-x)x(1+x)}{x^2+(1-x)a^2} \, \text{d}x=-\frac{1}{2} -a^2 -\left(1 +a^2 -a^4\right)\ln(a) -\left(\frac{a^{5}-3a^{3}-a}{ \sqrt{4 -a^2}}\right)\text{arccot}\left(\frac{a}{\sqrt{4-a^2}}\right)} $$ And since the last summand goes to $0$ for small $a$, you indeed verify that the integral $\sim - \frac{1}{2}-\ln(a) $ as $a \to 0^+$.