# Suppose that ax^2 + bx + c is a quadratic polynomial and that the integration: \int...

## Question:

Suppose that {eq}ax^2 + bx + c {/eq} is a quadratic polynomial and that the integration:

{eq}\int \frac{1}{ax^2 + bx + c} \text{d}x {/eq},

produces a function with neither a logarithmic nor inverse tangent term. What does this tell you about the roots of the polynomial?

## Integrals as Antiderivatives:

If we want to find the indefinite integral of a function, we are looking for its antiderivative. We have a variety of defined antiderivatives, and a few different methods for which to find these antiderivatives, including substitution and partial fraction decomposition.

If the integration of this rational function does not yield a logarithmic or inverse tangent function, then we know our integrand can't be of specific forms. This is because the basic forms of these antiderivatives come from the following definitions.

{eq}\begin{align*} \int \frac{1}{x\pm a} dx = \frac{1}{x\pm a} + c \\ \int \frac{1}{1+x^2} = \tan^{-1} x + c \end{align*} {/eq}

If our integrand is of the form {eq}\frac{1}{ax^2 + bx + c} {/eq}, then the way to integrate is by attempting to factor this denominator and if necessary perform partial fraction decomposition. When trying to factor a quadratic expression, we could arrive at a finite amount of possible expressions. This is because a quadratic polynomial will either have two distinct real roots, a repeated root, or two complex roots.

First, if the quadratic polynomial has two distinct real roots, which we will call d and e, then we can express this polynomial the following way using partial fraction decomposition.

{eq}\begin{align*} \frac{1}{ax^2 + bx + c} &= \frac{\frac{1}{a}}{x^2 + \frac{b}{a} x + \frac{c}{a}}\\ &= \frac{A}{x-d} + \frac{B}{x-e} \end{align*} {/eq}

However, these two terms can be integrated by applying the first definition, which yields a logarithmic expression. Thus, we know that the roots cannot be two real numbers, since the integration did not yield a logarithmic expression.

Second, if the polynomial has a repeated root, which we'll call d, then we can express this polynomial the following way. We can then integrate it by performing substitution.

{eq}\begin{align*} \frac{1}{ax^2 + bx + c} &= \frac{\frac{1}{a}}{x^2 + \frac{b}{a} x + \frac{c}{a}}\\ &= \frac{\frac{1}{a}}{(x-d)^2} \\ \int \frac{\frac{1}{a}}{(x-d)^2} dx &= \frac{1}{a} \int \frac{1}{u^2}\\ &= \frac{1}{a} \cdot -\frac{1}{u} + c\\ &= -\frac{1}{a(x-d)} + c \end{align*} {/eq}

This is neither a logarithmic nor an inverse trigonometric expression, so it's possible that this polynomial had one repeated root.

The final case is if the polynomial has two complex roots. In this case, we can't factor it directly. Instead, we need to complete the square, giving us an expression such as {eq}(x-d)^2 + e {/eq} and perform u-substitution where {eq}u = x-d {/eq}. If this new expression is factorable, which happens if {eq}e < 0 {/eq}, then we can factor this new denominator into two linear factors, perform partial fraction decomposition, and be in the same situation as the first one we described. If this new expression is not factorable, which happens if {eq}e > 0 {/eq}, then we need to manipulate this expression slightly to turn it into one of the form {eq}\frac{1}{\sqrt e }\cdot \frac{1}{(\frac{x}{\sqrt e})^2+1} {/eq}. The antiderivative of this expression is an inverse tangent function. Either way, two complex roots yields either a logarithmic or inverse tangent antiderivative, which means that this function can't have two real roots.

Therefore, this polynomial must have one repeated root.