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August 29, 2014

August 29, 2014

Posted by **mathstudent** on Thursday, January 10, 2008 at 2:41pm.

Int 1/(ax^2 + bx + c) dx

produces a function with neither a logarithmic or inverse tangent term. What does this tell you about the roots of the polynomial?

- Calculus -
**Damon**, Thursday, January 10, 2008 at 3:43pmwell int dx/[x(ax+b)] = 1/b log x/(ax+b)

if c were zero, that is what we would have, and x = 0 would be a root.

So if x = 0 is a root, no good, we get a log.

then int dx/(p^2+x^2) = (1/p)tan^-1 x/p

so we do not want b = 0 either

so we do not want roots of form

x= +/- sqrt (c/a)

That is all I can think of off hand.

- Calculus -
**Count Iblis**, Thursday, January 10, 2008 at 6:36pmFirst, to simplify things, observe that arctan can be expressed as a logarithm in terms of complex numbers. Suppose we have two different roots (complex or real), y1 and y2. Then:

ax^2 + bx + c = A(x-y1)(x-y2)

for some contant A

We have:

1/[(x-y1)(x-y2)] =

p[1/(x-y1) - 1/(x-y2)]

with

p = 1/(y1-y2)

So, the integral is then clearly a logarithm which can be written as an arctan if the roots are complex. Now any second degree polynomial has two roots in the set of complex numbers, however, the two roots can coincide. If that happens then the polynomial is proportional to:

1/(x-y1)^2

If we integrate this we obtain a term proportional to 1/(x-y1), which is not a logarithm nor an arctan. So, the only way to avoid a logarithmic or arctan term is if the two roots coincide to form a single root (we say that the root has a multiplicity of 2, when counting roots it counts double).

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