# Indefinite Integral of $n$-th power of Quadratic Denominator

I'd like to compute the indefinite integral

$$\int \frac{dx}{(ax^2+bx+c)^n}, a,b,c \in \mathbb{R}$$

without resorting to the usual recursion relation method of solution

i.e. without using integration by parts.

But I'd also like to do it without simplifying $ax^2+bx+c$ into anything like $t^2+1$, because if I have to remember to use that simplification then I might as well just memorize the solution itself. I'd like to just work it out from first principles instead.

There are two possible methods.

a) differentiating with respect to a parameter (Hardy's method):

If I differentiate $\frac{1}{ax^2+bx+c}$, with respect to $c$, $n-1$ times we have

$${\frac { \left( -1 \right) ^{n-1} }{ \left( n-1 \right) !}}{\frac {\partial }{\partial c^{n-1}}}\frac{1}{ a{x}^{2}+bx+c } = \frac{1}{ \left(a{x}^{2}+bx+c \right)^n}$$

Then, if we know

$$\int \! \frac{1}{ a{x}^{2}+bx+c }{dx}=2\,\arctan \left( { \frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c a-{b}^{2}}}}$$

the answer is simply

$$\int \! \frac{1}{ \left(a{x}^{2}+bx+c \right)^n}{dx}= {\frac { \left( -1 \right) ^{n-1} }{ \left( n-1 \right) !}}{\frac {\partial }{\partial c^{n-1}}} \left( 2\,\arctan \left( {\frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,ca-{b}^{2}}}} \right)$$

But computing $\int \! \frac{1}{ a{x}^{2}+bx+c }{dx}=2\,\arctan \left( { \frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c a-{b}^{2}}}}$ requires using those ugly $t^2+1$ simplifications, so we need another method.

b) Partial fractions allowing for complex roots

For n = 1, we have

$$\int \frac{dx}{ax^2+bx+c}=\frac{1}{a}\int \frac{dx}{(x-\alpha)(x-\beta)}=\frac{1}{a} \int \left( \frac{ \frac{1}{\alpha-\beta} }{x-\alpha} - \frac{\frac{1}{\alpha-\beta}}{x-\beta}\right)dx=\frac{1}{a} \frac{1}{\alpha-\beta} \ln\left|\frac{x-\alpha}{x-\beta}\right|$$

But the usual solution is given as

$$\int \! \frac{1}{ a{x}^{2}+bx+c }{dx}=2\,\arctan \left( { \frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c a-{b}^{2}}}}$$

This raises the question: is

$$\frac{1}{a} \frac{1}{\alpha-\beta} \ln\left|\frac{x-\alpha}{x-\beta}\right| = 2\,\arctan \left( { \frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c a-{b}^{2}}}}$$

true for both cases, i.e. when the quadratic has either two real, non-zero, roots or complex conjugate roots?

If it is true, could someone show me how to derive the arctan form from the logarithm form, in both real root and complex conjugate root cases?

(I looked in Bronstein's Symbolic Integration book and it looks like a very complicated matter to show these are equal, but I may be misunderstanding this)

Once the n = 1 case is solved, I believe the n > 1 case will be solved because

but this method does not look like it will give the arctan recursion relation solution

So, can someone help me integrate $\int \frac{dx}{(ax^2+bx+c)^n}, a,b,c \in \mathbb{R}$?

• I think this was asked yesterday... – JP McCarthy May 28 '14 at 16:43