We call $f(x) = \frac{p(x)}{q(x)}$ a rational function if both $p(x)$ and $q(x)$ are polynomials. If the degree of $p(x)$ is less than the degree of $q(x)$, then we call $f(x)$ a proper rational function

Distinct Linear Factors

1. If we want to integrate $\frac{2x^2+37x+144}{x^2+13x+40}$, we need to first perform long division to get $2 + \frac{11x+64}{x^2+13x+40}$

2. Now we need to integrate $\frac{11x+64}{x^2+13x+40}$ by splitting it up into distinct linear factors

3. If we factor the denominator, we get $\frac{11x+64}{(x+5)(x+8)}$

   1. We have two distinct linear factors, $x+5$ and $x+8$

   <aside> 👉 We can always find numbers $A$ and $B$ to make the following split (decomposition) work: $\frac{11x+64}{(x+5)(x+8)} = \frac{A}{x+5} + \frac{B}{x+8}$

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4. Now we need to get the coefficients $A$ and $B$ by multiplying by the common denominator, $11x+64 =A(x+8)+B(x+5)$

5. Rearranging this by powers of $x$, we get $11x+64 = (A+B)x + (8A+5B)$

6. Comparing powers of $x$, we now have two separate equations that we can solve: $A+B =11$ and $8A+5B = 64$

   1. $A = 3$ and $B=8$

7. Now we can integrate the original function

   

Distinct Quadratic Factors

If there are no real solutions to $ax^2 + bx+c = 0$, then it is irreducible. We need a polynomial of degree 1 in the numerator for an irreducible quadratic factor

1. Let’s try to solve $\int \frac{1}{x(x^2+1)}dx$

2. Given the polynomial $x^2+1$, we cannot write it as a product of linear factors

3. We need to write $\frac{1}{x(x^2+1} = \frac{A}{x}+ \frac{Bx+C}{x^2+1)}$

4. Solving for $A,B$ and $C$, we get $A= 1$, $B=-1$, and $C=0$

5. Now we can integrate the original function

   

Overview

When we want to simplify a rational function, we…

1. Use long division to get a proper rational function
2. Split the denominator in a product of linear factors and/or irreducible quadratic factors
   1. Always double check that a qaudratic factor really doesn’t have roots
   2. If it has a root, you must split it further into linear factors first