Integration by partial fractions Concepts

[ClipsClips99]

Hi, I am a little stuck on this conceptual question regarding integration by partial fractions.

Suppose that ax^2 + bx + c
is a quadratic polynomial and that the integration
int_ 1/ (ax^2 + bx + c )
produces a function with neither logarthmic nor inverse tangent terms. What does this tell you about the roots of the polynomial?

Lets see: If the integration doesnt produce a function with inverse tangent terms. I think that the roots must be real and that they can be factored out and put into partial fractions. But what about the logarthmic?

 

[skeeter]

ax² + bx + c either has two real roots or no real roots.

if the two real roots are distinct, then ax² + bx + c is factorable into two distinct linear factors and partial fraction decomposition will yield two terms that, when integrated, will yield a log function.

if ax² + bx + c has one real root of multiplicity two, then it will have the form (dx + e)², and the integral of 1/(dx+e)² will not be a log or arctan function.

if ax^2 + bx + c has no real roots, then the integral of 1/(ax² + bx + c) will yield an arctan function.

 

[soroban]

Hello, ClipsClips99!

An interesting problem . . .

Suppose that \(\displaystyle ax^2\,+\,bx\,+\,c\) is a quadratic polynomial
and that the integration: \(\displaystyle \L\,\int\frac{dx}{ax^2\,+\,bx\,+\,c}\) produces a function
. . with neither logarthmic nor inverse tangent terms.
What does this tell you about the roots of the polynomial?

Lets see: If the integration doesnt produce a function with inverse tangent terms,
I think that the roots must be real and that it can be factored and put into partial fractions.
But what about the logarthmic?

If we integrate the partial fractions, we get a logarithmic answer.

Example: \(\displaystyle \L\:\int\frac{dx}{(x\,-\,3)(x\,-\,2)} \;=\;\int\left(\frac{1}{x\,-\,3}\,-\,\frac{1}{x\,-\,2}\right)\)

. . \(\displaystyle \L= \;\ln|x\,-\,3|\,-\,\ln|x\,-\,2|\,+\,C \;=\; \ln\left|\frac{x\,-\,3}{x\,-\,2}\right|\,+\,C\)


It seems that if the roots are real, we get logs
. . and if the roots are complex, we get inverse tangent.
So what's left??


After I calmed down, I realized that this is a Discriminant problem
. . and there are three cases to consider.

If \(\displaystyle D\,>\,0\), the roots are real and distinct.
If \(\displaystyle D\,<\,0\), the roots are complex conjugates.
. . And if \(\displaystyle D\text{ equals }0\), the roots are real and equal.

In that case, the quadratic factors into the form: \(\displaystyle \,a(x\,+\,k)^2\)
. . and the integral is: \(\displaystyle \L\:\frac{1}{a}\int(x\,+\,k)^{-2}dx\;=\;-\frac{1}{a}(x\,+\,k)^{-1}\,+\,C\)
. . which has neither logs nor arctangent.


Answer: This tells us that the roots are equal.


Ha! .Skeeter beat me to it!