# Thread: Difficult hypothetical integral question

1. ## Difficult hypothetical integral question

Hi there-

I have a tricky question on a math worksheet that's phrased as a hypothetical problem. Basically, it says that if I take the integral of the reciprocal of a quadratic polynomial; if it produces no arctan terms, what does this say about the roots of the quadratic?

It then furthers the question and asks the same but if it produces no arctan or logarithmic terms.

I have no clue where to start with the first part, but I have noticed that reciprocals of quadratics with a double root [such as 1/(x-1)^2 ] do not produce arctan or log terms when you integrate them.

2. Originally Posted by reverendpeggy
Hi there-

I have a tricky question on a math worksheet that's phrased as a hypothetical problem. Basically, it says that if I take the integral of the reciprocal of a quadratic polynomial; if it produces no arctan terms, what does this say about the roots of the quadratic?

It then furthers the question and asks the same but if it produces no arctan or logarithmic terms.

I have no clue where to start with the first part, but I have noticed that reciprocals of quadratics with a double root [such as 1/(x-1)^2 ] do not produce arctan or log terms when you integrate them.
I would first note that in the grand scheme of things arctan and logarithmic terms really mean the same thing, it depends on whether there are only real roots for the denominator or are also complex roots [not factorizable in the real number system]. To see what I am talking about, look at
http://www.wolframalpha.com/input/?i...E2+%2B+25))+dx
where there are 2 complex double roots. The integral is only arctans [note that 4 is the square of 16 and 5 is the square root of 25]. Now change one of the plus' to a minus, say x2+16 to x2-16. The arctan is still there for the one factor but the one involving the x2-16 has changed to a logarithmic term. Since
x2-16 = x2+(-16)
we have the arctan and logarithmic terms are the same in the grand scheme of things, i.e. in our example replace 4=$\sqrt{16}$ by the complex 4i = $\sqrt{-16}$. Or, for the problem just note that if roots [of irreducible factors both of second order] are real you get logs and if roots are complex you get arctans.

One other thing I would note is that if we have complex roots they must occur in conjugate pairs, i.e if c=a+ib is a root so is c*=a-ib where i is the square root of -1 and a and b are real. If we multiply this pair together we get a quadratic equation with no real roots
(x-c) * (x-c*) = x2 + 2a x + a2 + b2

For the answer to the problem, I don't know anyway to do the problem other than just work through it. I would start with partial decomposition. [I'll do the 'easy' one for you noting that we always have four roots, some of which may be complex (see above about complex roots)]:
First we can replace our quadratic by (if necessary) a unit coefficient for the x4 term, i.e. consider
16 x4 + 32 x3 + 16 x2 = 16 [ x4 + 2 x3 + x2 ]
So let
f(x) = x4 + a3 x3 + a2 x2 + a1 x + a0
1) No multiple roots
Using partial decomposition one gets f(x) is a sum of four inverse linear functions with roots cj, i.e. $\frac{b_j}{x\, -\, c_j},\,\, j=1,\, 2,\, 3,\, 4; c_j\ne c_k\,\,\, if\, j\ne\, k$
Thus the answer is a series of logrithmic [or arctan depending on whether we have complex roots for f(x)] functions.

2) One multiple root
...

Happy hunting

3. Originally Posted by Denis
Welcome back Ishuda....nobody missed you
I figured but then I found out i missed the corner. I can't just go there so i figured if i tried to answer enough questions i would make a mistake and get to visit.

4. Originally Posted by Ishuda
I figured but then I found out i missed the corner. I can't just go there so i figured if i tried to answer enough questions i would make a mistake and get to visit.
And visit Denis ... at his favorite spot...