evaluate $\int_{0}^{1}dx \frac{(1-x)x(1+x)}{x^2+(1-x)a^2}$ I have been trying to solve the following integral:
$$\int_{0}^{1}dx \frac{(1-x)x(1+x)}{x^2+(1-x)a^2}$$
So far, I can see that when $a \rightarrow 0$, the integral diverges logarithmically.
Now, I want to evaluate it in the limit of small non-zero $a$. I expect that:
$$\int_{0}^{1}dx \frac{(1-x)x(1+x)}{x^2+(1-x)a^2} \sim C_0 + C_1 \log(a^2)$$ But then I'm stuck. How can I compute $C_0$ and $C_1$ ? i.e. how can I compute the integral for small $a$?
 A: 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 $a \to 0^+$.
A: Note
\begin{align}
\int_{0}^{1}\frac{(1-x)x(1+x)}{x^2+(1-x)a^2}dx
=& \int_{0}^{1}\frac{(1-x^2)x}{(x^2+a^2)(1-\frac{xa^2}{x^2+a^2})}dx
\approx 
 \int_{0}^{1}\frac{(1-x^2)x}{x^2+a^2}dx\\
=&\>\frac12 (1+a^2)\ln (\frac1{a^2}+1)-\frac12\overset{a\to 0}= -\frac12\ln a^2 -\frac12
\end{align}
