Integrals of inverse factored quadratic I am looking for solutions to integrals of the inverse of factored quadratic polynomials, i.e. $$\int \left( (x - K_1)(x-K_2 ) \right)^{-1} dx$$
There are probably look-up tables out there, but I would like to study a proof.
 A: $\dfrac{1}{(x-a)(x-b)}=\dfrac{1}{b-a}\left(\dfrac{1}{x-b}-\dfrac{1}{x-a}\right)$
Now integrating this you have ,
$\displaystyle \dfrac{1}{b-a}\ln\left(\dfrac{x-b}{x-a}\right)+C$
A: Here is a proof for the case you're interested in. I change a bit the notation.
Consider the polynomial in $x$ given by $P(x)=x^2-(a+b)x+ab$, which factors as $P(x)=(x-a)(x-b)$.
You look for constants $\alpha$ and $\beta$ such that ${1\over P(x)}={1\over (x-a)(x-b)}={\alpha\over x-a}+{\beta\over x-b}$: this condition gives you that ${(\alpha+\beta)x-\alpha b-\beta a\over (x-a)(x-b)}={1\over (x-a)(x-b)}$, which yields the system (in the unknowns $\alpha,\beta$)
$$
\left\{\begin{array}{c}
\alpha+\beta=0\\
\alpha b+\beta a=-1
\end{array}\right.
$$
Then $\beta=-\alpha$ and $\alpha(b-a)=-1$, so $\alpha=-\beta=-{1\over b-a}$. Thus ${1\over P(x)}=-{1\over b-a}\left({1\over x-a}+{1\over x-b}\right)$.
Therefore, you have:
$$
\int{1\over P(x)}\mathrm{d}x=\int\left({\alpha\over x-a}+{\beta\over x-b}\right)\mathrm{d}x=-{1\over b-a}\int{1\over x-a}\mathrm{d}x+{1\over b-a}\int{1\over x-b}\mathrm{d}x,
$$
which I assume you can solve.
