Q:

Extracting the divergent part of an integral

I want to evaluate the integral
$$ \int_0^1 \frac{2x(x-2)(1-x)}{(1-x)^2 + ax} \, \mathrm{d}x$$
in the limit of small $a$. For $a = 0$ this integral is divergent due to the $1/(1-x)$ pole. The exact expression for this integral is, according to Mathematica, a horrible combination of logs and inverse hyperbolic functions. However, I don't really care about all of the terms that go to zero as $a$ goes to zero. In, other words, I want to deduce that this integral is
$$1 + \ln a + \mathcal{O}(a)$$
(I believe this is the correct answer) by making some approximation at an early stage that saves me from having to compute the entire integral and then take the small $a$ limit.

A:

The integral can be rewritten explicitely as
$$
-2\int_0^1 \frac{x(x^2-3x+2)}{x^2-(2-a)x+1}\mathrm{d}x=-2 \int_0^1 x\left(1-\frac{(a+1)x-1}{x^2-(2-a)x+1}\right)\mathrm{d}x,
$$
hence it is equal to
$$
-1+2\int_0^1 \frac{(a+1)x^2-x}{x^2-(2-a)x+1}\mathrm{d}x=-1+2(a+1)+2\int_0^1 \frac{(-a^2+a+1)x-(a+1)}{x^2-(2-a)x+1}\mathrm{d}x
$$
$$
=1+2a+(-a^2+a+1)\int_0^1 \frac{2x-\frac{2(a+1)}{-a^2+a+1}}{x^2-(2-a)x+1}\mathrm{d}x
$$
$$
=1+2a+(-a^2+a+1)\int_0^1 \frac{2x-(2-a)}{x^2-(2-a)x+1}\mathrm{d}x+(-a^2+a+1)\left(2-a-\frac{2(a+1)}{-a^2+a+1}\right)\int_0^1\frac{1}{x^2-(2-a)x+1}\mathrm{d}x .
$$
We conclude that it is equal to
$$
1+2a+(-a^2+a+1)\ln|a|+(-a^2+a+1)\left(2-a-\frac{2(a+1)}{-a^2+a+1}\right)\int_0^1\frac{1}{x^2-(2-a)x+1}\mathrm{d}x.  
$$
You can calculate it explicitely, anyway it already proves your approximation
$$
1+\ln a+\mathcal{O}(a)
$$
holds true when $a \to 0^+$.

