Context: Oxford Concise Dictionary of Mathematics by Richard Earl and James Nicholson, 6th Edition (2021).
Self-study as a hobbyist.
We are given:
For $I_n (a, b) := \displaystyle \int \dfrac {a x + b} {(x^2 + A x + B)^n} \mathrm d x$ where $n \ge 2$, then:
$$I_n (a, b) = \dfrac {b A – 2 a B + (2 b – a A) x} {(n – 1) (4 B – A^2) (x^2 + A x + B)^n} + \dfrac {(2 n – 3) (2 b – a A)} {(n – 1) (4 B – A^2)} I_{n – 1} (0, 1)$$
My mission is to try to prove this.
I have used a similar technique to that used on this question:
Reduction formula for $I_{n, k} = \int x^k \left({x^2 + A x + B}\right)^n \ \mathrm d x$
as follows.
Let $h$ be the real function defined as:
$$\forall x \in \mathbb R: h (x) = x^2 + A x + B$$
Thus we have:
$$I_n (a, b) := \int \dfrac {a x + b} {(h (x))^n} \mathrm d x$$
Then, in order to get the suggestive coefficients in the answer given above, we do:
$$\dfrac {\mathrm d} {\mathrm d x} \left({\dfrac {a x + b} {(h (x))^{n – 1} } }\right) = (a x + b) \dfrac {-(n – 1)} {(h (x))^n} (2 x + A) + \dfrac a {(h (x))^{n – 1} }$$
using the product rule, chain rule and power rules for diffing.
After considerable algebra, we reach:
$$\dfrac {\mathrm d} {\mathrm d x} \left({\dfrac {a x + b} {(h (x))^{n – 1} } }\right) = -(n – 1) \dfrac {b A – 2 a B + (2 b – a A) x} {(h (x))^n} + \dfrac {a (3 – 2 n)} {(h (x))^{n – 1} }$$
The idea is then that we integrate both sides wrt $x$ to get:
$$\dfrac {a x + b} {(h (x))^{n – 1} } + a (2 n – 3) \int \dfrac {\mathrm d x} {(h (x))^{n – 1} } = -(n – 1) \int \dfrac {b A – 2 a B + (2 b – a A) x} {(h (x))^n} \mathrm d x$$
Expressing in terms of $I_n$ this is:
$$\dfrac {a x + b} {(x^2 + A x + B)^{n – 1} } + a (2 n – 3) I_{n – 1} (0, 1) = -(n – 1) (b A – 2 a B + (2 b – a A)) I_n (1, 0)$$
which is nowhere near where we are aiming.
We note that we have that required $I_{n – 1} (0, 1)$, and the coefficients $(n – 1$ and $(2 n – 3)$ in ''almost'' the right places, but even after we bend and strain the result by piling on the algebra to force $I_n (a, b)$ to appear, the fact remains that can't seem to get the term $\displaystyle \int \dfrac {a x + b} {(h (n))^n} \mathrm d x$ without leaving a term in $\displaystyle \int \dfrac Q {(h (n))^n} \mathrm d x$ where $Q$ is messy.
Nothing else I've tried comes remotely close to anything even approaching the expected coefficients, and nothing seems to allow me to eliminate spurious terms of $I_n$ and/or $I_{n – 1}$.
Anyone have any ideas as to what my approach might need to be?
Best Answer
With $h(x)=x^2+Ax+B$ and $K_n=\int \frac1{h(x)^n}dx$, establish $$\left(\frac{2x+A}{h(x)^{n-1}} \right)’= \frac{2(3-2n)}{h(x)^{n-1}}+\frac{(n-1)(4B-A^2)}{h(x)^{n}} $$ and, in turn $$\frac{2x+A}{h(x)^{n-1}}= 2(3-2n)K_{n-1}+(n-1)(4B-A^2)K_n\tag1 $$ Note that
\begin{align} I_n (a, b) &= \int \dfrac {a x + b} {(x^2 + A x + B)^n} d x\\ &= \frac a2\int \dfrac {2 x + A} {h(x)^n} d x + \frac {2 b -aA}2\int \frac {1} {h(x)^n} d x\\ &= -\frac a{2(n-1)h(x)^{n-1}}+ \frac {2 b -aA}2K_n \end{align} Utilize (1) to replace $K_n$, along with $K_{n-1}=I_{n-1}(0,1)$
$$I_n (a, b) = \dfrac {b A - 2 a B + (2 b - a A) x} {(n - 1) (4 B - A^2) h(x)^{n-1}} + \dfrac {(2 n - 3) (2 b - a A)} {(n - 1) (4 B - A^2)} I_{n - 1} (0, 1)$$