Partial Fraction method

This covers only a part (p.366 ~ 368) of Section 7.4 because the rest will not be on the exam.

Fun fact. I once tried to find some new facts behind this method during my undergraduate years, but pretty much everything I could imagine was known to the math community. However, I got more interest in abstract algebra because of this method.

Starting example. Consider computing the following indefinite integral: $$\int \frac{2x + 3}{(x + 1)(x + 2)}dx.$$ At first, you might freeze, but once you FOIL the bottom, you get $$\int \frac{2x + 3}{(x + 1)(x + 2)}dx = \int \frac{2x + 3}{x^{2} + 3x + 2}dx.$$ Now, you might notice that you could take $u = x^{2} + 3x + 2$ so that $du = (2x + 3)dx.$ This lets us compute $$\begin{align*}\int \frac{2x + 3}{(x + 1)(x + 2)}dx &= \int \frac{2x + 3}{x^{2} + 3x + 2}dx \\ &= \int \frac{1}{u}du \\ &=  \ln(|u|) + c \\ &= \ln(|x^{2} + 3x + 2|) + c,\end{align*}$$ where $c$ is an arbitrary constant.

Let's tweak the problem. What if you need to compute $$\int \frac{1}{(x + 1)(x + 2)}dx$$ instead? Our previous trick does not work well, so we need find a different method. Well, note that $$\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{(x+2) - (x+1)}{(x+1)(x+2)} = \frac{1}{(x + 1)(x + 2)}.$$ What's so big deal? The right-hand side was our integrand, and the left-hand side has something we can integrate! Thus, we have $$\int \frac{1}{(x + 1)(x + 2)} dx = \int \frac{1}{x + 1} - \frac{1}{x + 2}dx = \ln(|x+1|) - \ln(|x+2|) + c,$$ where $c$ is the integrating constant.

A more systematic method (a.k.a Partial Fraction method). How could we find such magical decomposition into partial fractions? Well, write $$\begin{align*}\frac{1}{(x + 1)(x + 2)} &= \frac{A}{x+1} + \frac{B}{x + 2} \\ &= \frac{A(x+2) + B(x+1)}{(x+1)(x+2)} \\ &= \frac{(A+B)x+(2A + B)}{(x+1)(x+2)}. \end{align*}$$ Comparing the numerators of the left-hand side and the right-hand side, we get $$0x + 1 = 1 = (A+B)x + (2A + B).$$ This means that

• $A + B = 0$ and
• $2A + B = 1.$

Subtracting the first equation from the second equation, we get $A = (2A + B) - (A + B) = 1 - 0 = 1.$ But then from the first equation, we see that $B = -1.$ This establishes $$\frac{1}{(x + 1)(x + 2)}  = \frac{A}{x+1} + \frac{B}{x + 2} = \frac{1}{x+1} - \frac{1}{x + 2}.$$ Let's do some exercises.

Exercise. Compute the following integral using the Partial Fraction method: $$\int \frac{2x + 3}{x^{2} + 3x + 2}dx.$$

Exercise. Compute $$\int \frac{2x + 3}{x^{2} - 3x + 2}dx.$$

Exercise. Compute $$\int \frac{2x + 3}{x^{2} + 3x + 2}dx.$$

Exercise (Hard). Compute $$\int \frac{2x + 3}{x^{2} + 2x + 2}dx.$$ (Hint: I don't know how to do this with the Partial Fraction method. I would first take care of $2x + 2$ much of the numerator using the substitution trick. For the rest, note that our integrand becomes $1/(1 + (x+1)^{2}) = \frac{d}{dx} \arctan(x + 1).$)

Exercise. Do #1 (c) on Exam 1 (Winter 2019).