The Integral $\int \frac {dx}{(x^2-2ax+b)^n}$ Recently I came across this general integral,
$$\int \frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±\sqrt {a^2-b} = a±\sqrt {∆}$$
Hence the integrand can be written as,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\frac {1}{(x-a-\sqrt ∆)^n(x-a+\sqrt ∆)^n}
$$
Resolving into partial fractions we have,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\sum \frac {A_r}{(x-a-\sqrt ∆)^r} + \sum \frac {B_r}{(x-a+\sqrt ∆)^r}
$$
Putting $-\frac {1}{2\sqrt ∆} = D$ , I could produce a table of the coefficients $A$ and $B$ for different $n$.
\par
For $n=1$,
$$A_1=-D , B_1=D$$
For $n=2$,
$$A_1=2D^3 , B_1=-2D^3$$
$$A_2=D^2 , B_2 = D^2$$
For $n=3$,
$$A_1=-6D^5 , B_1=6D^5$$
$$A_2=-3D^4 , B_2 = -3D^4$$
$$A_3=-D^3, B_3=D^3$$
For $n=4$,
$$A_1=20D^7, B_1=-20D^7$$
$$A_2=10D^6 , B_2 = 10D^6$$
$$A_3=4D^5, B_3=-4D^5$$
$$A_4=D^4, B_4=D^4$$
For $n=5$,
$$A_1=-70D^9, B_1=70D^9$$
$$A_2=-35D^8, B_2 = -35D^8$$
$$A_3=-15D^7, B_3=15D^7$$
$$A_4=-5D^6, B_4=-5D^6$$
$$A_5=-D^5, B_4=D^5$$
Yet I am unable to deduce a general formula for the coefficients. If I have the coefficients, the integral is almost solved , for then I shall have a logarithmic term and a rational function in $x$. More directly, I seek a result of the form,
$$\kappa \log \left( \frac {x-a-\sqrt ∆}{x-a+\sqrt ∆}\right) + \frac {P(x)}{Q(x)}$$
Any help would be greatly appreciated.
Conjecture 1(Proved below)
$$A(n,r)= (-1)^n \binom {2n-r-1}{n-1} D^{2n-r}$$
$$B(n,r)= (-1)^{n-r} \binom {2n-r-1}{n-1} D^{2n-r}$$

Let $b\neq a^2$,
$$S(n)=\int \frac {dx}{(x^2-2ax+b)^n}$$
With method of  undetermined coefficients we find formula
$$S(n)=\frac{Ax+B}{(x^2-2ax+b)^{n-1}}+CS(n-1)$$
We get
$$1=-\left( 2 A n-C-3 A\right) \, {{x}^{2}}-\left( \left( 2 B-2 A a\right)  n+\left( 2 C+4 A\right)  a-2 B\right)  x\\+2 B a n-\left( -C-A\right)  b-2 B a$$
$$A=\frac{1}{2 \left( b-{{a}^{2}}\right) \, \left( n-1\right) },B=-\frac{a}{2 \left( b-{{a}^{2}}\right) \, \left( n-1\right) },\\C=\frac{2 n-3}{2 \left( b-{{a}^{2}}\right) \, \left( n-1\right) }$$
Then
$$S(n)=\frac{x-a}{2(n-1)(b-a^2)(x^2-2ax+b)^{n-1}}+
\frac{2n-3}{2(n-1)(b-a^2)}S(n-1), \; n>1$$
$$S(1)=\int \frac {dx}{x^2-2ax+b}$$
