# Integral with quadratic denominator^n

1. May 28, 2014

### integral_int

1. The problem statement, all variables and given/known data

I'd like to compute

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

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 $t^2+1$, 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.

2. Relevant equations

There are two possible methods.

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

b) Partial fractions allowing for complex roots

3. The attempt at a solution

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

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2. May 28, 2014

### SammyS

Staff Emeritus
So, you don't like the integration by parts which gives "the usual recursion relation".

Then you have trouble with a solution which -- " this method does not look like it will give the arctan recursion relation solution".

That is all very puzzling .

I suggest integration by parts & evaluating the n = 1 case using completing the square.

3. May 29, 2014

### integral_int

Thanks for the response.

I don't understand what's puzzling about what I've asked help with? The partial fractions method should be equivalent to the recursion relation method, and in fact they are according to Bronstein. However I cannot see why

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

If I follow your hint and complete the square on the denominator for n = 1 I am basically using a $t^2+1p$ simplification, and I'd like to avoid that method. Why integrate a quadratic denominator which gets hairy, when we can just straight-forwardly integrate linear denominators then convert it into the answer we'd get had we integrated the quadratic denominator?