Finding the integral of the type $\frac{px+q}{ax^2 +bx + c}$

Question

The textbook says, to find the integral of the type $\dfrac{px+q}{ax^2 +bx + c}$, where $p,q,a,b,c$ are constants, we are to find real numbers $A$ and $B$ such that

$$px+q = A \dfrac{d}{dx} (ax^2 + bx + c) + B => A(2ax+b) + B.$$

Now to determine $A$ and $B$, we equate both sides of the coefficients of $x$ and constant terms so the integral is reduced to one of the known forms [such as "$\dfrac{1}{x^2 - a^2}$"], and then we can find the integral easily.

But, Can you please explain why we have to differentiate the denominator of the given integral? I am not able to see how it works. Why do we have to find $\frac{d}{dx}$ of $(ax^2 + bx + c)$? How does it work out?

Thank you

Answer 1

Let $f/g$ be your fraction. If you can rewrite $f$ in the form $f = a g' + b$ for constants $a, b$, then $$\int \frac{f}{g} = \int \frac{ag'+b}{g} = a \int \frac{g'}{g} + b \int \frac{1}{g},$$ and you may directly integrate $\frac{g'}{g}$ as $\ln |g|$. This reduces the problem to computing an integral of the form $\int 1/g$ where $g$ is a quadratic polynomial, so either one of the standard primitives $$ \int \frac{dx}{x^2}, \quad\int \frac{dx}{x^2+a^2} \quad\text{or} \int \frac{dx}{x^2-a^2}.$$ (Depending on the discriminant of your quadratic function).

Answer 2

If you can find $A$ and $B$ satisfying the required conditions, then integral of given type can be rewritten $$\int \left(\frac{A(2ax+b)}{ax^2+bx+c} + \frac{B}{ax^2+bx+c}\right)dx$$ Differentiating the denominator is what enables us to write the integral in this form. Then the integral of the function on the left can be found easily using substitution and the integral of the function on the right can be found by completing the square and making another substitution.

Here's how you can find $A$ and $B.$ Firstly, that arrow you have in your question should really be an equals sign. Then $px+q=A(2ax+b)+B.$ Equate coefficients of $x$ on the left and right and do the same for the constant terms (much like you would for the method of partial fractions). This sets up a system of 2 equations in $A$ and $B$ which you can solve to find $A$ and $B$ in terms of $p,q,a$ and $b.$