1
$\begingroup$

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.

2 Answers 2

3

$\dfrac{1}{(x-a)(x-b)}=\dfrac{1}{b-a}\left(\dfrac{1}{x-b}-\dfrac{1}{x-a}\right)$

Now integrating this you have ,

$\displaystyle \dfrac{1}{b-a}\ln\left(\dfrac{x-b}{x-a}\right)+C$

  • 0
    The answer involves the ln? Hmm. How do we manage the situations where $x = b$ or $x = a$? – 2017-02-26
  • 0
    It won't converge if any of the limits are either a or b – 2017-02-26
0

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.