# 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)

$$A(n,r)= (-1)^n \binom {2n-r-1}{n-1} D^{2n-r}$$ $$B(n,r)= (-1)^{n-r} \binom {2n-r-1}{n-1} D^{2n-r}$$

• The notation $\;\Delta\;$ is usually connected to the discriminant of a quadratic equation. In your case, it'd be $\;x_{1,2}=a\pm\sqrt\Delta\;$...which leads to the second comment: what if $\;\Delta<0\;$ ? Then you have two complex non-real roots... Jan 25, 2019 at 8:46
• The decomposition still makes sense for negative discriminant - it'll just lead to an answer in complex form. Jan 25, 2019 at 8:49
• @DonAntonio, As for the notation I have amended it, as for the sign of the discriminant , in case it is negative the logarithmic term with imaginary argument can be written as inverse sine function. Jan 25, 2019 at 9:15
• $\frac{A_k}{B_k} = (-1)^k$ for $k = 1,...,n$ except when n = 3. Are the signs right in that case for k = 2, 3?
– Paul
Jan 25, 2019 at 9:18
• A conjecture: $A(n,k)=(-1)^n\binom{2n-k-1}{n-1}D^{2n-k}$ and $B(n,k)=(-1)^{n-k}\binom{2n-k-1}{n-1}D^{2n-k}$. It fits the data given (with the corrected signs). No full solution yet, and I feel like moving on to something else for a while. Jan 25, 2019 at 9:46

All right, now I've got it.

The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting $$z=x-a-\sqrt{\Delta}$$ and later defining $$D=\frac1{2\sqrt{\Delta}}$$, we get \begin{align*}\frac1{(x^2-2ax+b)^n} &= \frac1{(x-a-\sqrt{\Delta})^n(x-a+\sqrt{\Delta})^n}=\frac1{z^n(z+2\sqrt{\Delta})^n}=\frac1{z^n}\cdot\frac{(2\sqrt{\Delta})^{-n}}{(1+\frac{z}{2\sqrt{\Delta}})^n}\\ \frac1{(x^2-2ax+b)^n} &= \frac{(-D)^n}{z^n(1-Dz)^n} = \frac{(-D)^n}{z^n}\sum_{j=0}^{\infty} \binom{n+j-1}{j}D^jz^j\\ &=(-1)^n\sum_{j=0}^{\infty}\binom{n+j-1}{j}D^{n+j}z^{j-n}\end{align*} We claim that the coefficients $$(-1)^n\binom{n+j-1}{j}D^{n+j}$$ for $$j are precisely the coefficients of $$\frac1{z^{n-j}}$$ in the partial fractions expansion of $$\frac1{z^n(z+2\sqrt{\delta})^n}$$. Why? Subtract the negative-exponent terms of the Laurent series from the partial fractions expansion. The difference is locally bounded, with a nice power series. But then, the only terms in the partial fractions expansion that aren't locally bounded are the $$\frac1{z^k}$$ terms - so their coefficients all have to match with the terms from the Laurent series.

Let $$k=n-j$$, and we get $$A(n,k)=(-1)^n\binom{2n-k-1}{n-k}D^{2n-k}=(-1)^n\binom{2n-k-1}{n-1}D^{2n-k}$$ in the partial fractions expansion $$\frac1{z^n(z+2\sqrt{\Delta})^n}=\sum_{k=1}^n \frac{A(n,k)}{z^k} +\sum_{k=1}^n \frac{B(n,k)}{(z+2\sqrt{\Delta})^k}=\sum_{k=1}^n \frac{A(n,k)}{(x-a-\sqrt{\Delta})^k} +\sum_{k=1}^n \frac{B(n,k)}{(x-a+\sqrt{\Delta})^k}$$ Oh, yes - in my comment, I didn't actually define my notation, and the update to the question imported that without defining it. The purpose is clear; we're just putting both parameters in the notation instead of just the power $$k$$ of $$\frac1{z-a\pm\sqrt{\Delta}}$$. Formally, the definition is the line just above.

That's half of the conjecture. For the other half, we expand around the other root. \begin{align*}\frac1{(x^2-2ax+b)^n} &= \frac1{(x-a-\sqrt{\Delta})^n(x-a+\sqrt{\Delta})^n}=\frac1{(w-2\sqrt{\Delta})^nw^n}=\frac1{w^n}\cdot\frac{(-2\sqrt{\Delta})^{-n}}{(1-\frac{w}{2\sqrt{\Delta}})^n}\\ \frac1{(x^2-2ax+b)^n} &= \frac{D^n}{w^n(1+Dw)^n} = \frac{D^n}{w^n}\sum_{j=0}^{\infty} \binom{n+j-1}{j}(-D)^jw^j\\ &=\sum_{j=0}^{\infty}(-1)^j\binom{n+j-1}{j}D^{n+j}w^{j-n}\end{align*} Again, extract the negative-exponent terms to get $$B(n,k)=(-1)^{n-k}\binom{2n-k-1}{n-k}D^{2n-k} =(-1)^{n-k}\binom{2n-k-1}{n-1}D^{2n-k}$$. The conjecture is confirmed, and we have our general formula.

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 $$S(n)=\frac{x-a}{2(n-1)(b-a^2)(x^2-2ax+b)^{n-1}}+ \frac{2n-3}{2(n-1)(b-a^2)}S(n-1), \; n>1$$ $$S(1)=\int \frac {dx}{x^2-2ax+b}$$

• Moreover, this recurrence can be solved in closed form, giving $$\int \frac {dx} {(x - a)^n (x - b)^n} = \frac {a + b - 2 x} {2 (2 n - 1) (x - a)^n (x - b)^n} \,{_2\hspace{-1px}F_1} {\left( 1, n; n + \frac 1 2; -\frac {(a - b)^2} {4 (x - a) (x - b)} \right)}.$$ Jan 25, 2019 at 14:56
• @Aleksas Somaras, though I did not mark your answer ,yet I appreciate your approach of using reduction formula and therefore it is my humble request that please don't remove your answer. Jan 26, 2019 at 4:12

Let $$a$$, $$b$$ and $$x$$ be real and $$n$$ be a positive integer. Let $$\Delta:= a^2-b$$. then the following formula holds: $$\begin{eqnarray} \frac{1}{(x^2-2 a x+b)^n} &=& \sum\limits_{l_1=1}^n \binom{2 n-1-l_1}{n-1} \frac{(-1)^n}{(x-a-\sqrt{\Delta})^{l_1}} \cdot \frac{1}{(-2 \sqrt{\Delta})^{2 n-l_1}} + \\ && \sum\limits_{l_1=1}^n \binom{2 n-1-l_1}{n-1} \frac{(-1)^n}{(x-a+\sqrt{\Delta})^{l_1}} \cdot \frac{1}{(+2 \sqrt{\Delta})^{2 n-l_1}} \end{eqnarray}$$ The result follows from the second formula from the top in my answer to How to quickly solve partial fractions equation? .