What is $\int_0^1 \frac{(1-x^2)xm^2}{m^2x^2+\lambda^2(1-x)} \, dx$? What is $$\int_0^1 \frac{(1-x^2)xm^2}{m^2x^2+\lambda^2(1-x)} \, dx \text{?}$$ I tried to compute it using Mathematica but it couldn't solve the integral. In fact, I am only interested in the part that diverges for $\lambda \rightarrow 0$, i.e., in terms up to $\mathcal{O(\lambda^0)}$.
 A: Let $\lambda = \gamma m,$ in order to remove the $m^2$s. $$ I(\gamma) := \int_0^1 \frac{(1-x^2) x}{x^2 + \gamma^2(1-x)}\,\mathrm{d}x \\ = \int_0^1 \frac{x -x^3}{(x - \gamma^2/2)^2 + (\gamma^2 - \gamma^4/4)}\,\mathrm{d}x.$$
Since we're interested in the behaviour for small $\gamma$, we may assume that $\gamma^2 > \gamma^4/4$. Let $\gamma^2 - \gamma^4/4 = \alpha^2$, and $x - \gamma^2/2 = u$. Also, let $\gamma^2/2 = \beta$ to ease notation. We have
$$\int_{-\beta}^{1-\beta} \frac{u + \beta - (u+\beta)^3}{u^2 + \alpha^2}\\ = \int_{-\beta}^{1-\beta} -u - 2\beta + \frac{(1-3\beta^2 - \alpha^2) u + \beta(1-3\alpha - \beta^2)}{u^2 + \alpha^2}$$
It is possible but tedious to exactly calculate the above. But we can see the behaviour as $\gamma \to 0$ more directly - notice that $\alpha = \gamma(1 + O(\gamma))$ and $\beta = \gamma^2/2$. Thus $\beta/\alpha = \gamma/2(1 + O(\gamma))$
Now, the first two integrals are easily seen to be $\int_{-\beta}^{1-\beta} - u = -\frac12 + O(\gamma)$, $\int - 2\beta = -2\beta = O(\gamma)$. Next note that $$ \int_{-\beta}^{1-\beta} \frac{\beta}{u^2 + \alpha^2} = \frac{\beta}{\alpha} \left(\arctan \frac{1- \beta}{\alpha} + \arctan \frac{\beta}{\alpha}\right).$$ Since $\arctan$ is bounded, and $\beta/\alpha = O(\gamma)$, this whole thing is $O(\gamma)$. This leaves us with $$ I(\gamma) = -\frac12 + O(\gamma) + \frac{(1-3\beta^2 - \alpha^2)}{2} \log \frac{(1-\beta)^2 + \alpha^2}{\beta^2 + \alpha^2}.$$ It's a matter of computation that $(1-\beta)^2 + \alpha^2 = 1$ and $\beta^2 + \alpha^2 = \gamma^2,$ giving us $$ I(\gamma) = \log(1/\gamma) - \frac{1}{2} + O(\gamma),$$ or, setting $\gamma = \lambda/m,$ giving us that the original integral behaves as $\log(1/\lambda) + \log(m) - \frac12 + O(\lambda)$.
